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3 years ago

6

A 1-kg ball is thrown at 10 m/s straight upward. neglecting air resistance, the net force that acts on the ball when it is half

way to the top of its path is about

2 answers:

Mama L [17]3 years ago

8 0

1kg of mass always has 9.8 Newtons of downward gravitational force acting on it, no matter where on Earth it is or how it's moving. And air resistance has no effect on the force.

Marrrta [24]3 years ago

4 0

Because of gravity it has 10N force downward

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The current theory of the structure of the

Mariana [72]

Answers:

a) $2.82(10)^{21} kg$

b) $1410 J$

c) $36.62 m/s$

Explanation:

<h3>a) Mass of the continent</h3>

Density $\rho$ is defined as a relation between mass $m$ and volume $V$ :

$\rho=\frac{m}{V}$ (1)

Where:

$\rho=2720 kg/m^{3}$ is the average density of the continent

$m$ is the mass of the continent

$V$ is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

$V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}$

Finding the mass:

$m=\rho V$ (2)

$m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3})$ (3)

$m=2.82(10)^{21} kg$ (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy $K$ is given by the following equation:

$K=\frac{1}{2}mv^{2}$ (5)

Where:

$m=2.82(10)^{21} kg$ is the mass of the continent

$v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s$ is the velocity of the continent

$K=\frac{1}{2}(2.82(10)^{21} kg)(1(10)^{-9} m/s)^{2}$ (6)

$K=1410 J$ (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass $m=77 kg$ and the same kinetic energy as that of the continent $1413 J$ , we can find its velocity by isolating $v$ from (5):

$v=\sqrt{\frac{2 K}{m}}$ (6)

$v=\sqrt{\frac{2 (1413 J)}{77 kg}}$

Finally:

$v=36.62 m/s$ This is the speed of the jogger

5 0

3 years ago

If i am changing velocity, i must also have _______ and a net _________

Alinara [238K]

if i am changing velocity, i must also have <u>acceleration</u> and a net <u>force</u>

<h2><u>Newton's</u><u> </u><u>first</u><u> </u><u>law</u><u> </u><u>of</u><u> </u><u>motio</u><u>n</u></h2>

Newton's first law of motion states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.

According to Newton's first law of motion, without a force acting on an object, its velocity does not change. The net force acts on an object to change its velocity and cause acceleration.

Read more about velocity:

brainly.com/question/4931057

6 0

2 years ago

Suppose you push a hockey puck of mass m across frictionless ice for a time \Delta t, starting from rest, giving thepuck speed v

bazaltina [42]

Answer:

1. $t_2 = 2t_1$

2. $t_2 = t_1\sqrt{2}$

Explanation:

1. According to Newton's law of motion, the puck motion is affected by the acceleration, which is generated by the push force F.

In Newton's 2nd law: F = ma

where m is the mass of the object and a is the resulted acceleration. So in the 2nd experiment, if we double the mass, a would be reduced by half.

$a_1 = 2a_2$

Since the puck start from rest, in the 1st experiment, to achieve speed of v it would take t time

$t = v / a_1$

Now that acceleration is halved:

$t = \frac{v}{2a_2}$

$\frac{v}{a_2} = 2t$

You would need to push for twice amount of time $t_2 = 2t_1$

2. The distance traveled by the puck is as the following equation:

$d = at^2$

So if the acceleration is halved while maintaining the same d:

$\frac{d_1}{d_2} = \frac{a_1t_1^2/2}{a_2t_2^2/2}$

As $d_1 = d_2$ , then $d_1/d_2 = 1$ . Also $a_1 = 2a_2$

$1 = \frac{2a_2t_1^2}{a_2t_2^2}$

$t_2^2 = 2t_1^2$

$t_2 = t_1\sqrt{2}\approx 1.14t_1$

So t increased by 1.14

7 0

3 years ago

What mass has a rest energy of 100J?

alexandr1967 [171]

Answer:

option a is correct

Explanation:

<h2>I hope it's help you ❣️❣️</h2>

6 0

2 years ago

How will the solubility of a gas solute change if the pressure above the solution is reduced?

Hoochie [10]

If the pressure above a solution containing a gas solute is reduced, the limit of the gas's solubility will decrease.

6 0

3 years ago

Read 2 more answers