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Citrus2011 [14]

3 years ago

11

Marcy pulls a backpack on wheels down the 100 m hall. The 60N force is applied at an angle of 30° above the horizontal. How much

work done on the weights? show full working out

1 answer:

shutvik [7]3 years ago

6 0

Work= Fcos∆×S

W=60N×cos 30⁰×100

W=60×0.866×100=5196.1J

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Can someone please help me out with this question thanks

IrinaK [193]

Answer:

1.98V

Explanation:

V =IR

V =Voltage

I =Current

R = Resistance $V =IR\\V = 100 (0.0198)\\V = 1.98 volts\\Voltage 1.98V$

8 0

3 years ago

If I was floating in space, why would a scale read my weight as "0"?

irina1246 [14]

Answer:

No external contact force.

Explanation:

Mass is defined as the amount of material an object contains, that doesn't change whether the object it on earth, on the moon or anywhere in space. Except the only force that is acting upon the body in gravity and they don't feel gravitational attraction towards the earth.

4 0

3 years ago

The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge

lesya [120]

Answer:

Loss, $\Delta E=-10.63\ J$

Explanation:

Given that,

Mass of particle 1, $m_1=m =0.66\ kg$

Mass of particle 2, $m_2=7.4m =4.884\ kg$

Speed of particle 1, $v_1=2v_o=2\times 6=12\ m/s$

Speed of particle 2, $v_2=v_o=6\ m/s$

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

$m_1v_1+m_2v_2=(m_1+m_2)V$

$V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}$

$V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}$

V = 6.71 m/s

Initial kinetic energy before the collision,

$K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)$

$K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)$

$K_i=135.43\ J$

Final kinetic energy after the collision,

$K_f=\dfrac{1}{2}(m_1+m_2)V^2$

$K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2$

$K_f=124.80\ J$

Lost in kinetic energy,

$\Delta K=K_f-K_i$

$\Delta K=124.80-135.43$

$\Delta E=-10.63\ J$

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

7 0

3 years ago

Suppose that the resistive force of the air on a skydiver can be approximated by f = −bv2. If the terminal velocity of an 82.0 k

antiseptic1488 [7]

Answer:

The value of b 0.7351 kg/m

Explanation:

Given that;

Mass of sky diver = 82 kg

Velocity = 33.4 m/s

f = −bv²

It is a resistance force, therefore the negative sign is ignored.

since; f = mg

∴ mg = bv²

b = mg / v ² ........ (1)

At terminal velocity a = 0

Put parameters in (1)

b = (82 × 10) / (33.4)²

b = 820 / 1,115.56

b = 0.7351

6 0

3 years ago

Mike and Mitchell decide to have a foot race. They mark off a stretch of 100 yards, and recruit Cindy to work the stopwatch. Aft

Vanyuwa [196]

Answer: A

Explanation: the lower the time, the faster you complete the race ( i also had this question before)

3 0

4 years ago