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lord [1]
2 years ago
5

If a bicyclist has a mass of 70 KG and a velocity of 25M/S, what is the momentum of the bicyclist?

Physics
1 answer:
ArbitrLikvidat [17]2 years ago
4 0

Answer:

1750 kgm/s

Explanation:

The equation for momentum is p = mv = 70 * 25 = 1750.

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What is the difference in KE between a 52.5 kg person running 3.50 m/s and a 0.0200 kg bullet flying 450 m/s?
rusak2 [61]

Answer:

Ek = 1705.28 [J]

Explanation:

In order to solve this problem, we must remember that kinetic energy can be calculated by means of the following equation.

E_{k}=\frac{1}{2} *m*v^{2}

where:

m = mass [kg]

v = velocity [m/s]

Ek = kinetic energy [J] (Units of Joules)

<u>For the person running</u>

<u />E_{k} =\frac{1}{2}*52.2*(3.5)^{2} \\ E_{k} =319.72[J]<u />

<u />

<u>For the bullet</u>

<u />E_{k} =\frac{1}{2} *m*v^{2}<u />

<u />E_{k} =\frac{1}{2} *0.02*(450)^{2} \\E_{k}=2025 [J]<u />

<u />

The difference in Kinetic energy is equal to:

Ek = 2025 - 319.72

Ek = 1705.28 [J]

8 0
2 years ago
Tyler wants to learn about the types of insects in the soil near his house. Which would be a benefit of carrying out a descripti
lara31 [8.8K]
To gather more information and details on the soil
8 0
2 years ago
Which technology contains at least one permanent magnet?
stich3 [128]
I’m pretty sure it’s, B. An Electric Motor. I apologize in advance if it’s incorrect.
6 0
2 years ago
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The third floor of a house is 8m above street level. How much work is needed to move a 136kg refrigerator to the third floor?
jonny [76]

m = Mass of the refrigerator to be moved to third floor = 136 kg

g = Acceleration due to gravity by earth on the refrigerator being moved = 9.8 m/s²

h = Height to which the refrigerator is moved  = 8 m

W = Work done in lifting the object

Work done in lifting the object is same as the gravitational potential energy gained by the refrigerator. hence

Work done = Gravitation potential energy of refrigerator

W = m g h

inserting the values

W = (136) (9.8) (8)

W = 10662.4 J



8 0
2 years ago
The speed of an arrow fired from a compound
san4es73 [151]

Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

<u>Equations</u>

X axis:

X=v_{ox}*t

v_{0x} =v_0cos(\alpha)

Y axis:

Y= Y_0 +v_{y0} t - \frac{g}{2} t^2

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, Y=0, then

0= Y_0 +v_{y0} t - \frac{g}{2} t^2

0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2

Caculate the segcond degree equation to obtain the two possible values for t:

t_1= 10.18 \\t_2= -0.04046

But, in physics, time it could not be negative, so we take t_1= 10.18

Caculate now:

X=79m/s*cos(\39)*10.18s= 624.996 m

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

v_0= 79m/s+13m/s= 92m/s

Using the same procedure that item A, caculate X

First, we need to know the new time

0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2

And we obtain:

t_1=11.845s\\t_2=-0.041s

One more time, we take the positive time: t_1=11.845s

Finally:

X=92m/s *cos(39)*11.845s=846.887 m

6 0
3 years ago
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