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lilavasa [31]
3 years ago
14

A ball resting on a roof 75 meters high has 1000 Joules of gravitational potential energy. Calculate the mass of the ball. (SHOW

ALL WORK)
Physics
1 answer:
Arturiano [62]3 years ago
8 0

Answer:

The mass of the ball is 1.360 kilograms.

Explanation:

By Work-Energy Theorem, gravitational potential energy (U), in joules, is the product of weight of the ball (W), in newtons, and height (h), in meters. Please notice that weight is the product of the mass of the ball (m) and gravitational acceleration (g), in meters per square second. Then, the formula for the mass of the ball is:

m = \frac{U}{g\cdot h} (1)

If we know that U = 1000\,J, g = 9.807\,\frac{m}{s^{2}} and h = 75\,m, then the mass of the ball is:

m = \frac{U}{g\cdot h}

m = 1.360\,kg

The mass of the ball is 1.360 kilograms.

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Two carts with masses of 4.2 kg and 3.2 kg move toward each other on a frictionless track with speeds of 5.4 m/s and 4.5 m/s, re
rodikova [14]

Answer:the final speed is 5.01 m/s

Explanation:

Momentum is the product of mass and velocity.

Cart 1 has a mass of 4.2kg and a speed 5.4 m/s

Cart 2 has a mass of 3.2kg and a speed 4.5 m/s

Total momentum before collision is

m1u1 + m2u2. It becomes

4.2×5.4 + 3.2×4.5 = 22.68 + 14.4

= 37.08kgm/s

The carts stick together after colliding head-on. This means that they move with a common velocity, v. Therefore, Total momentum after collision is (m1 + m2)v. It becomes

(4.2 + 3.2)v = 7.4v

According the the law of conservation of momentum, the total momentum before collision = the total momentum after collision. Therefore,

7.4v = 37.08

v = 37.08/7.4 = 5.01 m/s

8 0
3 years ago
A fixed 14.1-cm-diameter wire coil is perpendicular to a magnetic field 0.52 T pointing up. In 0.28 s , the field is changed to
ELEN [110]

Answer:

the average induced emf in the coil is 0.016 V.

Explanation:

Given;

diameter of the wire, d = 14.1 cm = 0.141 m

change in magnetic filed strength, dB = 0.52 T - 0.23 T = 0.29 T

change in time, dt = 0.28 s

The area of the wire is calculated as follows;

A = \frac{\pi d^2}{4} \\\\A = \frac{\pi \times (0.141)^2}{4} \\\\A = 0.0156 \ m ^2

The induced emf is calculated as follows;

emf = \frac{dBA}{dt} \\\\emf = \frac{0.29 \times 0.0156}{0.28} \\\\emf = 0.016 \ V

Therefore, the average induced emf in the coil is 0.016 V.

8 0
3 years ago
A revolutionary war cannon, with a mass of 2260 kg, fires a 15.5 kg ball horizontally. The cannonball has a speed of 109 m/s aft
Anton [14]

Answer:

The gain in velocity is 0.37m/s

Explanation:

We need solve this problem though the conservation of momentum. That is,

m_1 v_1 = m_2 v_2

m_1=2260Kg\\m_2=15.5Kg\\v_2= 109m/s

Using the equation to find v_1,

v_1=\frac{m_2 v_2}{m_1}\\v_1=\frac{15.5*109}{2260}\\v_1= 0.7475

Using the conservation of energy equation, we have,

KE= \frac{1}{2}m*v^2

KE_{cball}=\frac{1}{2}(15.5)(109)^2=92077.75J

KE_{cannon}=\frac{1}{2}(2260)(0.7475)^2=631.39J

Total KE= 92077.75+13425530=92708.9J

Now this energy over the cannonball

KE=\frac{1}{2}m*v_2^2

92708.9=\frac{1}{2}15.5v_2^2

V_2 = 109.37m/s

The gain in velocity is 0.37m/s

4 0
3 years ago
Please help me!!!!!!!!!
Nadusha1986 [10]

Hi there! :)

\large\boxed{17.32 m/s}

Use the following equation in solving for kinetic energy:

KE = 1/2mv² where:

KE = kinetic energy (J)

m = mass (kg)

v = velocity (m/s)

Plug in the given values:

12,000 = 40v²

Divide both sides by 40:

12,000 / 40 = v²

300 = v²

Take the square root of both sides:

√300 = v

v ≈ 17.32 m/s

4 0
3 years ago
An archer shoots an arrow toward a 300-g target that is sliding in her direction at a speed of 2.30 m/s on a smooth, slippery su
Xelga [282]

Answer:

<em>4.79m/s</em>

Explanation:

According to law of conservation of momentum;

The sum of momentum of the bodies before collision is equal to the momentum after collision.

m1u1 + m2u2 = (m1+m2)v

Given;

m1 = 0.3kg

u1 = 2.30m/s

m2 = 0.0225kg

u2 = 38m/s

Required

speed of the arrow after passing through the target v

Substituting the given data into the formula

0.3(2.3) + 0.0225(38) = (0.3 + 0.0225)v

0.69 + 0.855 = 0.3225v

1.545 = 0.3225v

v = 1.545/0.3225

v = 4.79m/s

<em>Hence the speed of the arrow after passing through the target is 4.79m/s</em>

3 0
3 years ago
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