A steel ball is dropped onto a thick piece of foam. The ball is released 2.5 meters above the foam. The foam compresses 3.0 cm as the ball comes to rest. What is the magnitude of the ball's acceleration as it comes to rest on the foam

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:

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].

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:

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:

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.

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:

ans: 2.25 meter

explanation

use following equations

F = ma

V = U + aT

S = UT + 1/2 aT^2

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].

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:

- 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

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]

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:

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.

Answer:

The answer should be clenching fists

Please delete my answer. I made a mistake

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. liquid and solid

Explanation:

Answer:

answer is d

Explanation:

i hope this helps you

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]

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:

we need the block

Explanation:

1×2 =4 lest 74 =345

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:

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

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

Explanation:

time=Distance/speed

t=32/8

t=4 seconds

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