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Sergeu [11.5K]
2 years ago
15

If the soccer player runs with a speed of 4.6m/s, how long does it take him to run 60m?

Physics
2 answers:
Diano4ka-milaya [45]2 years ago
7 0
About 13 s (60/4.6)
satela [25.4K]2 years ago
5 0

Explanation:

Hello there!

Just divide 60 by 4.6:

60/4.6=13.04 seconds.

;)

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Determine the sign (+ or −) of the torque about the elbow caused by the biceps, τbiceps, the sign of the weight of the forearm,
Alex Ar [27]
Ans: 
1.  τbiceps = +(Positive)
2.  τforearm = -(Negative)
3.  τball = -(Negative)

Explanation:

The figure is attached down below.

1. T<span>orque about the elbow caused by the biceps, τbiceps:
Since Torque = r x F (where r and F are the vectors)
</span>Where r is the vector from elbow to the biceps.
<span>
We can see in the figure that F(biceps) is in upward direction, and by applying the right hand rule from r to F, we get the counterclockwise direction. The torque in counterclockwise direction is positive(+). Therefore, the sign would be +.

2. </span>Torque about the the weight of the forearm, τforearm:
Since Torque = r x F (where r and F are the vectors)
Where r is the vector from elbow to the forearm.

Also weight is the special kind of Force caused by the gravity.

We can see in the figure that W(forearm) is in downward direction, and by applying the right hand rule from r to F, we get the clockwise direction. The torque in clockwise direction is negative(-). Therefore, the sign would be -.

3. Torque about the the weight of the ball, τball:
Since Torque = r x F (where r and F are the vectors)
Where r is the vector from elbow to the ball.

Also weight is the special kind of Force caused by the gravity.

We can see in the figure that W(ball) is in downward direction, and by applying the right hand rule from r to F, we get the clockwise direction. The torque in clockwise direction is negative(-). Therefore, the sign would be -.

8 0
2 years ago
How much gravitational potential energy does a 45.2 kg object have when it is 21.9m above the ground?
Blizzard [7]

Answer:

Explanation:

The formula for gravitational potential energy is

Ep = m · g · h   Assuming that the acceleration is g = 10m/s²

Ep = 45.4 · 10 · 21.9 = 9,942.6 J

God is with you!!!

6 0
1 year ago
Two billiard balls move toward each other on a table. The mass of the number three ball, m1, is 5 g with a velocity of 3 m/s. Th
Stels [109]

This question deals with the law of conservation of momentum, which basically says that the total momentum in a system must stay the same, provided there are no outside forces. Since you were given the mass and velocity of the two objects you can find the momentum (p=mv) of each and then add them together to find the total momentum of the system before they collide. This total momentum must be the same after they collide.  Since you have the mass and velocity of one of the objects after the collision you can find the its momentum after.  Subtract this from the the system total and you will have the momentum of the other object after the collision.  Now that you know the momentum of the other object you can find its velocity using p=mv and its mass from before.

Be careful with the velocities.  They are vectors, so direction matters.  Typically moving to the right is positive (+) and moving to the left is negative (-).  It is not clear from your question which direction the objects are moving before and after the collision.

6 0
2 years ago
Read 2 more answers
A small child gives a plastic frog a big push at the bottom of a slippery 2.0 meter long, 1.0 meter high ramp, starting it with
valentinak56 [21]
Refer to the diagram shown below.

Because the ramp is slippery, ignore dynamic friction.
Let m =  the mass of the frog.
g = 9.8 m/s²

The KE (kinetic energy) at the bottom of the ramp is
KE₁ = (1/2)*(m kg)*(5 m/s)² = 12.5 m J

Let v =  the velocity at the top of the ramp.
The KE at the top of the ramp is
KE₂ = (1/2)*m*v²= 0.5 mv² J
The PE (potential energy) at the top of the ramp relative to the bottom is
PE₂ = (m kg)*(9.8 m/s²)*(1 m) = 9.8m J

Conservation of energy requires that
KE₁ = KE₂ + PE₂
12.5m = 0.5mv² + 9.8m
0.5v² = 2.7
v = 2.324 m/s

Answer: 2.324 m/s

7 0
2 years ago
An isolated charged point particle produces an electric field with magnitude E at a point 2 m away. At a point 1 m from the part
guajiro [1.7K]

Explanation:

The electric field at a distance r from the charged particle is given by :

E=\dfrac{kq}{r^2}

k is electrostatic constant

if r = 2 m, electric field is given by :

E_1=\dfrac{kq}{(2)^2}\\\\=\dfrac{kq}{4}\ .....(1)

If r = 1 m, electric field is given by :

E_2=\dfrac{kq}{r_2^2}\\\\=\dfrac{kq}{1}\ ....(2)

Dividing equation (1) and (2) we get :

\dfrac{E_1}{E_2}=\dfrac{\dfrac{kq}{4}}{kq}\\\\\dfrac{E_1}{E_2}=\dfrac{1}{4}\\\\E_2=4\times E_1

So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.

4 0
1 year ago
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