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3 years ago

A boy rides a sled down a steep, snow.covered hill. Which of

Physics

2 answers:

shtirl [24]3 years ago

5 0

Answer:

Potential energy decreases and kinetic energy increases.

Explanation:

That was the answer on my CK12

Travka [436]3 years ago

4 0

Answer:

a. Potential energy decreases and Kinetic energy increases

Explanation:

Because as he comes down due to its steepness the speed of the boy or you can say his KE increases and since he comes from a high position (hill) to the lower ground his potential energy decreases simultaneously

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A 70 kg human sprinter can accelerate from rest to 10 m/s in 3.0 s. During the same time interval, a 30 kg greyhound can go from

ladessa [460]

Answer:

$P_1 = 1166.7 Watt$

$P_2 = 2000 Watt$

Explanation:

Average power for the human sprinter is given as

$Power = \frac{\Delta E}{\Delta t}$

so we have

$P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}$

$P = \frac{\frac{1}{2}(70)(10^2) - 0}{3}$

$P_1 = 1166.7 Watt$

Average power for greyhound is given as

$P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}$

$P = \frac{\frac{1}{2}(30)(20^2) - 0}{3}$

$P_2 = 2000 Watt$

3 0

3 years ago

A woman walked 115 m. As she did so, her speed increased from 4.20 m/s to 5.00 m/s. How long did it take her to walk this distan

ASHA 777 [7]

Answer:

25 seconds

Explanation:

Assuming the woman is accelerating at a constant rate of $a \;\;m/s^2$ from the initial velocity, u=4.20 m/s, to the final velocity, v=5.00 m/s.

Let she takes t seconds to cover the distance, s=115 m.

As acceleration, $a=\frac{v-u}{t}=\frac{5-4.2}{t}$

$\Rightarrow at=0.8\cdots(i)$

Now, from the equation of motion

$s=ut+\frac 12 at^2$

$\Rightarrow s=ut+\frac 12 at(t)$

$\Rightarrow 115=4.2t+\frac 12 \times 0.8 t$ [ from equation (i)]

$\Rightarrow 115=(4.2+0.4)t$

$\Rightarrow t= 115/4.6 = 25$ seconds.

Hence, she takes 25 seconds to walk the distance.

5 0

3 years ago

If the current through a 20-Ω resistor is 8.0 A , how much energy is dissipated by the resistor in 1.0 h ? Express your answer w

marshall27 [118]

Answer:

$P(3600)=593.247W$

Explanation:

First, let's find the voltage through the resistor using ohm's law:

$V=IR=20*8=160V$

AC power as function of time can be calculated as:

$P(t)= V*I*cos(\phi)-V*I*cos(2 \omega t-\phi)$ (1)

Where:

$\phi=Phase\hspace{3}angle\\\omega= Angular\hspace{3}frequency$

Because of the problem doesn't give us additional information, let's assume:

$\phi=0\\\omega=2 \pi f=2*\pi *(60)=120\pi$

Evaluating the equation (1) in t=3600 (Because 1h equal to 3600s):

$P(3600)=160*8*cos(0)-160*8*cos(2*120\pi*3600-0)\\P(3600)=1280-1280*cos(2714336.053)\\P(3600)=1280-1280*0.5365255751\\P(3600)=1280-686.7527361=593.2472639\approx=593.247W$

5 0

4 years ago

A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the

lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

$l^2 = x^2 + y^2-----(1)$