What is the current in the 30 resistor?
A. 0.0833 A
B. 12 A
C. 2 A
D. 10 A

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  

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

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

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]

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

Answer:

ans: 2.25 meter

explanation

use following equations

F = ma

V = U + aT

S = UT + 1/2 aT^2

Please delete my answer. I made a mistake

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:

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].

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

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

Answer:

4.02×10⁻⁴ K⁻¹

Explanation:

Applying,

γ = (v₂-v₁)/(v₁Δt)................. Equation 1

Where γ  = coefficient of volume expansion, v₂ = Final volume, v₁ = initial volume, Δt = change in temperature.

From the question,

Given: v₂ = 850 m³, v₁ = 830 m³, Δt = (90-30) = 60°C

Substitute these values into equation 1

γ  = (850-830)/(830×60)

γ  = 20/(830×60)

γ  = 4.02×10⁻⁴ K⁻¹

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]

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

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.

Answer:

hey. i dont know what you tryna say but if u replying to someone else, you should use the comments section. in that way you won't lose points.

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

Answer:

answer is d

Explanation:

i hope this helps you

Answer:

kinetic

Explanation:

i just remember it from last year

Answer:

kinetic energy

Explanation:

expression for kinetic energy is

kinetic energy = (1/2) × mass × (velocity)^2

so , as velocity increases K.E increases

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].

Explanation:

time=Distance/speed

t=32/8

t=4 seconds

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]

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.

Answer:

A. liquid and solid

Explanation:

Answer:pp

Explanation:

ii

1) (40 N) cos(30°) ≈ 34.6 N

2) (40 N) sin(30°) = 20 N

3) (65 kg) g = (65 kg) (9.80 m/s²) = 585 N

4) The net force on the sled acting in the vertical direction is made up of

• the sled's weight, 585 N, pointing downward

• the vertical component of the applied force, 20 N, pointing upward

• the normal force, with magnitude n, also pointing upward

The sled does not move up or down, so by Newton's second law,

∑ F = n + 20 N - 585 N = 0   ==>   n = 565 N

5) The net force in the horizontal direction consists of

• the horizontal component of the applied force, 34.6 N, acting in the direction the sled's movement (call this the positive direction)

• kinetic friction, with magnitude f, pointing in the opposite and negative direction

By Newton's second law,

∑ F = 34.6 N - f = 0   ==>   f ≈ 34.6 N

Now if µ is the coefficient of kinetic friction, then

f = µn   ==>   µ = f/n = (34.6 N) / (565 N) ≈ 0.0613

The component of the force is the effective part of that force in that direction.

The component of the force is the effective part of that force in that direction.

1) The horizontal component of a force = 40 N cos 30 degrees = 34.6 N

2) The vertical component of the force = 40 N sin 30 degrees = 20 N

3) The magnitude of the gravitational force = mg cos 30 degrees  = 65 Kg * 9.8 m/s^2 * cos 30 degrees = 551.7 N

4) The normal force = 551.7 N

5) The coefficient of friction = F/R =  40 N /551.7 N = 0.07

Learn more about component of a force:https://brainly.com/question/15529350

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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

Answer:

we need the block

Explanation:

1×2 =4 lest 74 =345