Answer:
the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²
Explanation:
Given the data in the question;
initial velocity; u = 0 m/s
height; h = 2.5 m
we find the velocity of the ball just before it touches the foam.
using the equation of motion;
v² = u² + 2gh
we know that acceleration due gravity g = 9.81 m/s²
so we substitute
v² = ( 0 )² + ( 2 × 9.81 × 2.5 )
v² = 49.05
v = √49.05
v = 7.00357 m/s
Now as the ball touches the foam
final velocity v₀ = 0 m/s
compresses S = 3 cm = 0.03 m
so
v₀² = v² + 2as
we substitute
( 0 )² = 49.05 + 0.06a
0.06a = -49.05
a = -49.05 / 0.06
a = -817.5 m/s²
Therefore, the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²
explanation on energy from air pressure light from water pressure
What must be true if energy is to be transferred as heat between two bodies in physical contact?
1-The two bodies must have different volumes.
2-The two bodies must be at different temperatures.
3-The two bodies must have different masses.
4-The two bodies must be in thermal equilibrium.
Answer:
answer is d
Explanation:
i hope this helps you
A professional boxer hits his opponent with a 1035 N horizontal blow that lasts 0.175 s. The opponent's total body mass is 120 kg and the blow strikes him near his center of mass and while he is motionless in midair. Determine the following.(a) The opponent's final velocity after the blow(b) Calculate the recoil velocity of the opponent's 5.0-kg head if hit in this manner, assuming the head does not initially transfer significant momentum to the boxer's body.
Answer:
(a) vf = 1.51 m/s
(b) vf = 36.22 m/s
Explanation:
The rate of change of momentum is equal to the force:
[tex]F = \frac{mv_f-mv_i}{t}[/tex]
[tex]Ft = m(v_f-v_i)[/tex]
where,
F = Force = 1035 N
t = time = 0.175 s
vi = initial speed = 0 m /s
vf = final speed = ?
(a)
m = mass of body = 120 kg
Therefore,
[tex](1035\ N)(0.175\ s)=(120\ kg)(v_f - 0\ m/s)\\\\v_f = \frac{181.125\ Ns}{120\ kg} \\\\[/tex]
vf = 1.51 m/s
(b)
m = mass of head = 5 kg
Therefore,
[tex](1035\ N)(0.175\ s)=(5\ kg)(v_f - 0\ m/s)\\\\v_f = \frac{181.125\ Ns}{5\ kg} \\\\[/tex]
vf = 36.22 m/s
Lightning can be studied with a Van de Graaff generator, which consists of a spherical dome on which charge is continuously deposited by a moving belt. Charge can be added until the electric field at the surface of the dome becomes equal to the dielectric strength of air. Any more charge leaks off in sparks. Assume the dome has a diameter of 25.0 cm and is surrounded by dry air with a "breakdown" electric field of 3.00 10^6 V/m.
Required:
a. What is the maximum potential of the dome?
b. What is the maximum charge on the dome?
Answer:
(a) V = 3.75 x 10^5 V
(b) q = 5.2 x 10^-6 C
Explanation:
Diameter, d = 25 cm
radius, r = 12.5 cm = 0.125 m
Electric field, E = 3 x 10^6 V/m
(a) The maximum potential is given by
[tex]V = E \times r \\\\V = 3\times 10^6\times 0.125\\\\V = 3.75\times10^5 V[/tex]
(b) The charge is given by
[tex]V = \frac{k q}{r}\\\\3.75\times10^5=\frac{9\times10^9\times q}{0.125}\\\\q = 5.2\times 10^{-6} C[/tex]
Write the prime factorization of 32. Use exponents when appropriate and order the factors
from least to greatest
A train mass of 2000kg and speed 35 m/s collides and sticks to an identical train that is initially at rest .After the collision (a) what is the final speed of the entangled system?
(b) what is the kinetitic energy of the system? compare the final kinetic energy to initial kinetic energy?
Answer:
The system would be moving at [tex]17.5\; \rm m \cdot s^{-1}[/tex].
The kinetic energy of this system would be [tex]612500\; \rm J \![/tex] after the collision.
[tex]612500\; \rm J[/tex] (same amount) of kinetic energy would be lost.
Explanation:
The momentum of an object is the product of its mass [tex]m[/tex] and its velocity [tex]v[/tex]. That is: [tex]p = m \cdot v[/tex].
Assume that external forces (e.g., friction) have no effect on this system. The total momentum of this system would stay the same before and after the collision.
Initial momentum of this system:
Moving train: [tex]\begin{aligned}p &= m \cdot v \\ &= 2000\; \rm kg \times 35\; \rm m \cdot s^{-1} \\ &= 70000\; \rm kg \cdot m \cdot s^{-1}\end{aligned}[/tex].Since the other train wasn't moving before the collision, its initial momentum would be [tex]0[/tex].Hence, the momentum of this system would be [tex]70000\; \rm kg \cdot m \cdot s^{-1}[/tex] before the collision.
Under the assumptions, the collision would not change the momentum of this system. Hence, the momentum of this system would continue to be [tex]70000\; \rm kg \cdot m \cdot s^{-1}[/tex] after the collision.
However, with two identical trains stuck to each other, the mass of this system would be twice that of just one train: [tex]m = 2 \times 2000\; \rm kg[/tex].
Calculate the new velocity of this system:
[tex]\begin{aligned} v &= \frac{p}{m}\\ &= \frac{70000\; \rm kg \cdot m \cdot s^{-1}}{2 \times 2000\; \rm kg} = 17.5\; \rm m\cdot s^{-1}\end{aligned}[/tex].
Calculate the kinetic energy of this system before and after the collision.
Before the collision:
[tex]\begin{aligned}& \text{KE(before)} \\ =\; & \text{KE(moving train)} + \text{KE(stationary train)}\\ =\; & \frac{1}{2} \, m(\text{one train}) \cdot (v(\text{moving train}))^{2} + 0 \\ = \; &\frac{1}{2} \times 2000 \times (35\; \rm m\cdot s^{-1})^{2} \\ = \; & 1225000\; \rm J \end{aligned}[/tex].
After the collision:
[tex]\begin{aligned}& \text{KE(after)} \\ =\; & \frac{1}{2} \, m(\text{two trains}) \cdot v^{2} \\ = \; &\frac{1}{2} \times (2\times 2000\; \rm kg) \times (17.5\; \rm m\cdot s^{-1})^{2} \\ = \; & 612500\; \rm J \end{aligned}[/tex].
Change to the kinetic energy of this system:
[tex]1225000\; \rm J - 612500\; \rm J = 612500\; \rm J[/tex].
Answer the following questions
1. Heat in liquid travels from
a) bottom to top
b) top to bottom
c) left to right
d) right to left
2. The direction of flow of heat is
a) always from a cooler body to a hotter body
b) always from a hotter body to cooler body
c) always from a body at a lower temperature to a body at a higher temperature
d) all the above
3. A cold steel spoon is dipped in a cup of hot milk. The steel spoon transfer the heat to its other end by the process of
a) convection
b) conduction
c) radiation
d) none of the above
Question: A car of mass 500kg travelling at 12m/s enters a stretch of road where there's a constant resistive force of 8000N. The car comes to a stop due to this resistive force. Calculate the distance travelled by the car before stopping.
Answer:
ans: 2.25 meter
explanation
use following equations
F = ma
V = U + aT
S = UT + 1/2 aT^2
please helpp!
convert 1N into dyne
In the given relation F=ma a stands for write there SI unit
Answer:
a. 1 Newton = 100000 Dyne
b. a represents acceleration.
Explanation:
Newton is the standard unit (S.I) of measurement of force. Converting 1 Newton to dyne we have;
1 Newton = 10⁵ Dyne
1 Newton = 100000 Dyne
Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.
Mathematically, it is given by the formula;
Force = mass * acceleration
[tex] F = ma[/tex]
Hence, we can deduce that a represents the acceleration of an object and it's measured in meters per seconds square.
Baseball runner with a mass of 70kg, moving at 2.7m/s and collides head-on into a shortstop with a mass of 85kg and a velocity of 1.6m/s. What will be the resultant velocity of the system when they make contact with each other
Answer:
The speed of the combined mass after the collision is 2.1 m/s.
Explanation:
mass of runner, m = 70 kg
speed of runner, u = 2.7 m/s
mass of shortstop, m' = 85 kg
speed of shortstop, u' = 1.6 m/s
Let the velocity of combined system is v.
Use conservation of momentum
Momentum before collision = momentum after collision
m u + m' u' = (m + m') v
70 x 2.7 + 85 x 1.6 = (70 + 85) v
189 + 136 = 155 v
v = 2.1 m/s
What is the torque in ( lbs-ft ) of a man pushing on a wrench with 65 lbs of force 8 unches from the nut / bolt he is trying to turn?
Explanation:
The torque [tex]\tau[/tex] is given by
[tex]\tau=Fd = (65\:\text{lbs})(\frac{8}{12}\:\text{ft}) = 43.3\:\text{lbs-ft}[/tex]
The ejection seat has an acceleration of 8gees (8xgravity or ~80m/s/s). He has a mass of 70kg. The total force on him from the chair/rocket would be ?
(80m/s/s)(70kg)=5600N
(80m/s/s)(70kg)=5600N + Fg = 5600N+(70kg)(9.8N/kg)~5600N+700N=6300N
(80m/s/s)(70kg)=5600N - Fg = 5600N+(70kg)(9.8N/kg)~5600N-700N=4900N
I need the time
please explain need this ASAP
I assume you're talking about a pilot. If the ejection seat has an acceleration of 8g, then it would exert a normal force of 8g (70 kg) ≈ 5600 N.
(This is assuming the pilot is flying horizontally at a constant speed, and the seat is ejected vertically upward.)
To reiterate, this is *only* the force exerted by the seat on the pilot. Contrast this with the net force on the pilot, which would be the normal force minus the pilot's weight, 5600 N - (70 kg)g ≈ 4900 N.
If instead the seat ejects the pilot directly downward, the force exerted by the seat would have the same magnitude of 5600 N, but its direction would be reversed to point downward, making it negative. But the net force would change to -5600 N - (70 kg)g ≈ -6300 N
A scooter is accelerated from rest at the rate of 8m/s
. How long will it take to cover
a distance of 32m?
Explanation:
time=Distance/speed
t=32/8
t=4 seconds
What is this sport ⚽⚾
Answer:
sports are all forms of physical activity that contribute to physical fitness, mental well-being and social interaction.
hope it is helpful to you
1. An AAMU basketball player is 2.03 meters tall. What is his height given in US customary units of feet and
inches?
Answer:
His height is 6.66 feet or 79.92 inches.
Explanation:
Given that,
An AAMU basketball player is 2.03 meters tall.
Let h is the height.
We know that,
1 m = 3.28 feet
So,
2.03 m = 6.66 feet
Also,
1 m = 39.37 inches
2.03 m = 79.92 inches
Hence, this is the required solution.
A bullet of mass 0.5 kg is moving horizontally with a speed of 50 m/s when it hits a block of mass 3 kg that is at rest on a horizontal surface with a coefficient of friction of 0.2. After the collision the bullet becomes embedded in the block. How much work is being dne by bullet?
Answer:
Work done by the bullet is 612.26 J.
Explanation:
mass of bullet, m = 0.5 kg
initial velocity of bullet, u = 50 m/s
coefficient of friction = 0.2
mass of block, M = 3 kg
let the final speed of the bullet block system is v.
use conservation of momentum
Momentum of bullet + momentum of block = momentum of bullet block system
0.5 x 50 + 3 x 0 = (3 + 0.5) v
v = 7.14 m/s
let the stopping distance is
The work done is given by change in kinetic energy of bullet
initial kinetic energy of bullet, K = 0.5 x 0.5 x 50 x 50 = 625 J
Final kinetic energy of bullet, K' = 0.5 x 0.5 x 7.14 x 7.14 = 12.74 J
So, the work done by the bullet
W = 625 - 12.74 = 612.26 J
You are driving home from school steadily at 97 km/h for 190 km . It then begins to rain and you slow to 60 km/h instantly. You arrive home after driving 4.0 hours.
how far is your hometown from school?
Please delete my answer. I made a mistake
Which hand position should be avoided in fitness walking?
flexing wrists
relaxing fingers
clenching fists
keeping hands loose
Answer:
The answer should be clenching fists
why acceleration independent variable
Answer:
Explanation:Force and acceleration are directly proportional. ... Mass and acceleration are inversely proportional. In this situation, acceleration changes in response to a change of mass, so mass is the independent variable and acceleration is the dependent variable.
Convert the following:
1) 367.5 mg = _______ g
2) 367 mL = _______ L
3) 28.59 in =______ cm
4) 8 0z =_______lb
5) 0.671 mm =_____m
Answer:
1) 0.3675
2) 0.367
3) 72.6186
4) 0.5
5) 0.000671
Answer:
1) 367.5 mg = 0.3675 g
2) 367 mL = 0.367 L
3) 28.59 in = 72.61 cm
4) 8 0z = 0.5 lb
5) 0.671 mm = 0.0000671 m
A total positive charge of 12.00 mC is evenly distributed on a straight thin rod of length 6.00 cm.
A positive point charge, Q = 4.00 nC, is located a distance of 5.00 cm above the midpoint of the
rod. What will be the electrical force on the point charge?
A 1500kg car is travelling at v=30m/s. The cars kinetic energy is? *
A) 45000J
B) 1350000J
C) 22500J
D)675000J
show your work please
Hi there!
[tex]\large\boxed{\text{D. 675000J}}[/tex]
Use the following formula to solve:
KE = 1/2mv², where:
KE = kinetic energy
m = mass (kg)
v = velocity (m/s)
Therefore:
KE = 1/2(1500)(30)²
KE = 1/2(1500)(900)
KE = 675000 J
Question 2:
Inclined Plane
A block (M) weighs 25-N, rests on an inclined plane when it is joined by a sting to a support
(S) as shown in the figure' below. Use g=10 N/Kg.
(S)
B
M
List and classify the forces acting on (M).
Représent, without scaling, the forces acting on (M).
Find the mass of (M).
74. If the string were cut, (M) does not slide. Explain this phenomenon.
15. Determine the mass and weight of (M) on moon.
06
Answer:
we need the block
Explanation:
1×2 =4 lest 74 =345
A car changes speed from 27m/s to 5m/s in 50m. The acceleration is: *
A) 7m/s2
B) 7.04m/s2
C) -7.04m/s2
D) 0.22m/s2
show your work please
by using v ^2 = u^2 + 2as we can find "a"
25 = 729 + 2 × a × 50
25 = 729 + 100a
a = - 7.04
so the answer is B
A 1200-kg car is being driven up a 5.0o hill. The frictional force is directed opposite to the motion of the car and has a magnitude of f = 524 N. A force F is applied to the car by the road and propels the car forward. In addition to these two forces, two other forces act on the car: its weight W and the normal force FN directed perpendicular to the road surface. The length of the road up the hill is 290 m. What should be the magnitude of F, so that the net work done by all the forces acting on the car is +150 kJ?
I suppose the hill makes an angle of 5.0° with the horizontal.
• F acts parallel to the road and in the direction of the car's motion, so it contributes a positive amount of work, F (290 m).
• Friction does negative work on the car since it opposes the car's motion. As the car moves up the slope, the work done by friction is (-524 N) (290 m) = -151,960 J.
• The car's weight has components that act parallel and perpendicular to the road. The parallel component has a magnitude of W sin(5.0°) and points down the slope, so it contributes negative work of -(1200 kg) g sin(5.0°) ≈ 1,024.95 J. The perpendicular component of W does not do any work.
• The normal force FN also doesn't do any work to move the car up the slope because it points perpendicular to the road, so we can ignore it, too.
The net work done on the car is then
F (290 m) + (-151,960 J) + 1,024.95 J = 150,000 J
==> F (290 m) ≈ 300,935 J
==> F ≈ (300,935 J) / (290 m) ≈ 1,037.71 N
In which states of matter will a substance have a fixed volume?
O A. Liquid and solid
O B. Solid and gas
O C. Plasma and gas
O D. Liquid and gas
Answer:
A. liquid and solid
Explanation:
Show that the speed with which a projectile leaves the ground is equal to its speed just before it strikes the ground at the end of its journey, assumilng the firing level equals the landing level.
Answer:
Thus, the velocity at the time of strike is same as the velocity at the time of projection.
Explanation:
Let a projectile is projected vertically upwards with a speed of u and reaches to the maximum height H.
At maximum height , the speed is zero and then the projective comes back on the ground.
Use the third equation of motion
[tex]v^2 = u^2 + 2 g h \\\\0 = u^2 - 2 g H\\\\\u =\sqrt{2gH}[/tex]
Now let the velocity at the time of strike is v'.
Use third equation of motion, here initial velocity is zero.
[tex]v'^2 = 0 + 2 g H \\\\v = \sqrt{2gH}[/tex]
Thus, the velocity at the time of strike is same as the velocity at the time of projection.
A 50 g copper calorimeter contains 250 g of water at 20 C. How much steam be condensed into the water to make the final temperature of the system 50 C. ( specific heat water= 4200 J/Kg C , specific heat copper= 390 J/Kg C
Answer:
Approximately [tex]13\; \rm g[/tex] of steam at [tex]100\; \rm ^\circ C[/tex] (assuming that the boiling point of water in this experiment is [tex]100\; \rm ^\circ C\![/tex].)
Explanation:
Latent heat of condensation/evaporation of water: [tex]2260\; \rm J \cdot g^{-1}[/tex].
Both mass values in this question are given in grams. Hence, convert the specific heat values from this question to [tex]\rm J \cdot g^{-1}[/tex].
Specific heat of water: [tex]4.2\; \rm J \cdot g^{-1}\cdot \rm K^{-1}[/tex].
Specific heat of copper: [tex]0.39\; \rm J \cdot g^{-1}\cdot K^{-1}[/tex].
The temperature of this calorimeter and the [tex]250\; \rm g[/tex] of water that it initially contains increased from [tex]20\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex]. Calculate the amount of energy that would be absorbed:
[tex]\begin{aligned}& Q(\text{copper}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 0.39\; \rm J \cdot g^{-1}\cdot K^{-1} \times 50\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 585\; \rm J \end{aligned}[/tex].
[tex]\begin{aligned}& Q(\text{cool water}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; \rm J \cdot g^{-1}\cdot K^{-1} \times 250\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 31500\; \rm J \end{aligned}[/tex].
Hence, it would take an extra [tex]585\; \rm J + 31500\; \rm J = 32085\; \rm J[/tex] of energy to increase the temperature of the calorimeter and the [tex]250\; \rm g[/tex] of water that it initially contains from [tex]20\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex].
Assume that it would take [tex]x[/tex] grams of steam at [tex]100\; \rm ^\circ C[/tex] ensure that the equilibrium temperature of the system is [tex]50\; \rm ^\circ C[/tex].
In other words, [tex]x\; \rm g[/tex] of steam at [tex]100\; \rm ^\circ C[/tex] would need to release [tex]32085\; \rm J[/tex] as it condenses (releases latent heat) and cools down to [tex]50\; \rm ^\circ C[/tex].
Latent heat of condensation from [tex]x\; \rm g[/tex] of steam: [tex]2260\; {\rm J \cdot g^{-1}} \times (x\; {\rm g}) = (2260\, x)\; \rm J[/tex].
Energy released when that [tex]x\; {\rm g}[/tex] of water from the steam cools down from [tex]100\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex]:
[tex]\begin{aligned}Q = \;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; {\rm J \cdot g^{-1}\cdot K^{-1}} \times (x\; \rm g) \times (100\;{\rm ^\circ C} - 50\;{\rm ^\circ C}) \\ =\; & (210\, x)\; \rm J \end{aligned}[/tex].
These two parts of energy should add up to [tex]32085\; \rm J[/tex]. That would be exactly what it would take to raise the temperature of the calorimeter and the water that it initially contains from [tex]20\; \rm ^\circ C[/tex] to [tex]50\; \rm ^\circ C[/tex].
[tex](2260\, x)\; {\rm J} + (210\, x)\; {\rm J} = 32085\; \rm J[/tex].
Solve for [tex]x[/tex]:
[tex]x \approx 13[/tex].
Hence, it would take approximately [tex]13\; \rm g[/tex] of steam at [tex]100\; \rm ^\circ C[/tex] for the equilibrium temperature of the system to be [tex]50\; \rm ^\circ C[/tex].
When rebuilding her car's engine, a physics major must exert 405 N of force to insert a dry steel piston into a steel cylinder. What is the magnitude of the normal force in newtons between the piston and cylinder
Answer:
[tex]N=675N[/tex]
Explanation:
From the question we are told that:
Force [tex]F=405N[/tex]
Generally the equation for Normal force in this case is is mathematically given by
[tex]F=\mu_s N[/tex]
Where
Static Friction=[tex]\mu_s[/tex]
[tex]\mu_s=0.6[/tex]
Therefore
[tex]N=\frac{F}{\mu_s}[/tex]
[tex]N=\frac{405}{0.6}[/tex]
[tex]N=675N[/tex]
A power plant generates 150 MW of electrical power. It uses a supply of 1000 MW from a geothermal source and rejects energy to the atmosphere. Find the power to the air and how much air should be flowed to the cooling tower (kg/s) if its temperature cannot be increased more than 10oC.
Answer:
- the power to the air is 850 MW
- mass flow rate of the air is 84577.11 kg/s
Explanation:
Given the data in the question;
Net power generated; [tex]W_{net[/tex] = 150 MW
Heat input; [tex]Q_k[/tex] = 1000 MW
Power to air = ?
For closed cycles
Power to air Q₀ = Heat input; [tex]Q_k[/tex] - Net power generated; [tex]W_{net[/tex]
we substitute
Power to air Q₀ = 1000 - 150
Q₀ = 850 MW
Therefore, the power to the air is 850 MW
given that ΔT = 10 °C
mass flow rate of air required will be;
⇒ Q₀ / CpΔT
we know that specific heat of air at p=c ; Cp = 1.005 kJ/kg.K
we substitute
⇒ ( 850 × 10³ ) / [ 1.005 × 10 ]
⇒ ( 850 × 10³ ) / 10.05
⇒ 84577.11 kg/s
Therefore, mass flow rate of the air is 84577.11 kg/s