A runner covers 40 m in 4.55 seconds ,what is their average velocity?

A weight lifter lifts a dumbbell a certain height in 2.0 s, while a competitor does the same workin 1.0 s. Compared to the power

Aloiza [94]

Answer:

a. one-half as great

Explanation:

The power developed by the first lifter is one-half as great as that of the second person.

Power is defined as the rate at which work is done;

Power = $\frac{workdone}{time}$

Since the two lifters do the same work at different time, let us estimate their power;

P₁ = $\frac{workdone}{2}$ P₂ = $\frac{workdone }{1}$

We see that for P₁, power is half of the work done whereas in P₂ power is the same as the work done.

Therefore,

The power of the first weight lifter is one-half the second lifter.

4 0

2 years ago

Where are alkaline earth metals found on the periodic table?

mojhsa [17]

Answer:

B

Explanation:

8 0

2 years ago

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

3 years ago

1. Which of the following transformations is an example of a β−-decay?

KatRina [158]

Answer:

1. C: 31/14 Si becomes 31/15 because a nuetron

2. A: 238 92U because the very long half-life means a very small rate of decay

3. D: Charge conservation is not satisfied

4. B: of the four nuclear decay processes only the α-decay changes the baryon number and does so in increments of four

Explanation:

I just took the quick check. Enjoy the answers I did not get to have

4 0

2 years ago

A golf ball reaches a height of 150 m before it stops rising and starts to fall to the ground. What is the golf balls speed (rou

artcher [175]

Answer:

v = 54 m/s

Explanation:

Given,

The maximum height of the flight of golf ball, h = 150 m

The velocity at height h, u = 0

The velocity of the golf ball right before it hits the ground, v = ?

Using the III equations of motion

Substituting the given values in the above equation,

v² = 0 + 2 x 9.8 x 150 m

= 2940

v = 54 m/s

Hence, the speed of the golf ball right before it hits the ground, v = 54 m/s

4 0

3 years ago