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2 years ago

7

A boat is floating in a small pond. the boat then sinks so that it is completely submerged. what happens to the level of the pon

d?

1 answer:

lukranit [14]2 years ago

4 0

A boat is floating in a small pond. the boat then sinks so that it is completely submerged. what happens to the level of the pond?
It increases!

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A weightlifter lifts a 13.0-kg barbel from the ground an moves it a distance of 1.3 meters. What is the work se does on the barb

marta [7]

The work done on the barbell is -165.62 Nm.

Explanation:

Work done on any object is the measure of force required to move that object from one position to another. So it is determined by the product of force acting on the object with the displacement of the object.

In the present problem, the displacement of the object on acting of force is given as 1.3 m. And the weight of the object which is a barbel is given as 13 kg. As the work is to lift the object from the ground, so the acceleration due to gravity will be acting on the object. In other words, the force applied on the object to lift it should be in opposite direction to the acting of acceleration due to gravity.

Thus, $Force = - Mass * Acceleration due to gravity = - 13 * 9.8 =-127.4 N$

Now, the force is -127.4 N and the displacement is 1.3 m.

So, $Work done = F*d$

$Work done = -127.4* 1.3 = -165.62 Nm$

So, the work done on the barbell is -165.62 Nm.

6 0

2 years ago

Mary takes 6.0 seconds to run up a flight of stairs that is 102 meters long. if mary's weight is 87 newtons, what power has mary

pentagon [3]

Thank you for posting your question here at brainly. I hope the answer will help. Below are the choices that can be found elsewhere:

<span>A. 1.5 * 10^3 Watts
B. 7.3 * 10^2 Watts
C. 3.5 * 10^2 Watts
D. 2.5 * 10^2 Watts
</span>
<span>Work = force*displacement = 10^2*87 = 8,700 joule
Power = work/time = 8,700/6 = 1.45*10^3 (rounded up to 1.5 kw). The answer is A. </span>

3 0

2 years ago

Read 2 more answers

Consider a bird that flies at an average speed of 10.7 m/sm/s and releases energy from its body fat reserves at an average rate

Wittaler [7]

Answer:

455165.278 m

Explanation:

P = Power = 3.7 W

v = Velocity = 10.7 m/s

Amount of fat = 4 g

1 gram of fat provides about 9.40 (food) Calories

Energy given by 4 g of fat

$E=4\times 9.4\times 4186\\\Rightarrow E=157393.6\ J$

Time required to burn the fat

$t=\dfrac{E}{P}\\\Rightarrow t=\dfrac{157393.6}{3.7}\\\Rightarrow t=42538.811\ s$

Distance traveled by the bird

$s=vt\\\Rightarrow s=10.7\times 42538.811\\\Rightarrow s=455165.2777\ m$

The bird will fly 455165.278 m

4 0

2 years ago

Read 2 more answers

You throw a tennis ball (mass 0.0570 kg) vertically upward. It leaves your hand moving at 15.0 m/s. Air resistance cannot be neg

Deffense [45]

Answer:195 J

Explanation:

Given

mass of ball $m=0.0570\ kg$

ball leaves the hand with $u=15\ m/s$

maximum height reached by ball $h=8\ m$

Initial Mechanical energy when ball just leaves the hand

$M.E._1=(P.E.+K.E.)_1$

$M.E._1=(mgh)_1+(\frac{1}{2}mv^2)_1$

considering hand to be datum so h_1=0[/tex]

so Potential energy at ground is zero

$M.E._1=\frac{1}{2}\times m\times (15)^2$

$M.E._1=6.41\ J$

Mechanical Energy at highest point

$(M.E.)_2=(P.E.+K.E.)_2$

at highest Point velocity is zero

$(M.E.)_2=mgh_2+0$

$(M.E.)_2=0.0570\times 9.8\times 8$

$(M.E.)_2=4.46\ J$

Decrease in Mechanical energy

$(M.E.)_1-(M.E.)_2=6.41-4.46$

$(M.E.)_1-(M.E.)_2=1.95\ J$

3 0

2 years ago

A boy throws a 15 kg ball at 4.7 m/s to a 65 kg girl who is stationary and standing on a skateboard. After catching the ball, th

jek_recluse [69]

Answer:

$a)v_{f}=0.88m/s$

Explanation:

To solve this problem we use the Momentum's conservation Law, before and after the girl catch the ball:

$\\ p_{1}=p_{2}\\m_{ball}*v_{o.ball}+m_{girl}*v_{o.girl} = m_{ball}*v_{f.ball} + m_{girl}*v_{f.girl}$ (1)

At the beginning the girl is stationary:

$v_{o.girl}=0m/s$ (2)

If the girl catch the ball, both have the same speed:

$v_{f.girl}=v_{f.ball}=v_{f}$ (3)

We replace (2) and (3) in (1):

$m_{ball}*v_{o.ball} = (m_{ball}+m_{girl})*v_{f} \\$

We can now solve the equation for v_{f}:

$v_{f}=\frac{m_{ball}*v_{o.ball}}{(m_{ball}+m_{girl})}=\frac{15*4.7}{15+65}=0.88m/s$

4 0

2 years ago