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zepelin [54]
3 years ago
8

A football coach sits on a sled while two of his players build their strength by dragging the sled across the field with ropes.

The friction fource on the sled is 1000N and the angle between the two ropes is 20°. How hard must each player pull to drag the coach at a steady 2.0m/s?
Physics
1 answer:
Jobisdone [24]3 years ago
5 0

Answer:

F=507.7N

Explanation:

According to Newton's second law:

\sum F=m.a

in this case, the football players need to drag the coach at a constant velocity, thus means with no acceleration, so:

\sum F=0\\\\-F_f+F_p*cos(\theta)=0\\\\

there are 20 degrees between the two ropes that means each player exerts a force 10 degrees from the zero reference.

F_p=2*F\\\\F=\frac{1000N}{2*cos(10)}\\\\F=507.7N

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the force of a bag of potatoes on a vegetable stand is 22.5 N down. what is the force of the vegetable stand on the bag of potat
natima [27]

As per Newton's III law we can say that

Force applied by object 1 on object 2 is always equal in magnitude and opposite in direction of the force that object 2 apply on object 1.

So we can say it as

F_{12} = -F_{21}

now here above question is based upon the same

if a bag of vegetables applied a force F = 22.5 N of the surface stand the the same surface will apply same magnitude of force in opposite direction on the vegetables bag

So our answer will be F = 22.5 N (upwards).

4 0
3 years ago
Which change increases the electric force between objects?
Brrunno [24]
Well, the force is proportional to the product of the charges
on the two objects.  So if the objects are already negatively
charged distance between them is unchanged, then adding
electrons to either or both objects would increase the forces
between them.

3 0
3 years ago
Read 2 more answers
Object A of mass 0.70 kg travels horizontally on a frictionless surface at 20 m/s. It collides with object B, which is initially
AnnZ [28]

Answer:

25.71 kgm/s

Explanation:

Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.

Given that K₂ = 0.7K₁

1/2mv₂² = 0.7(1/2mv₁²)

v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s

Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.

Th magnitude of object A's momentum change is thus 25.71 kgm/s

6 0
3 years ago
A physics student throws a softball straight up into the air. The ball was in the air for a total of 3.56 s before it was caught
meriva

Answer:

The initial velocity of the softball is 14.711 meters per second.

Explanation:

This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.

From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:

y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2} (Eq. 1)

Where:

y_{o} - Initial height of the softball, measured in meters.

y - Final height of the softball, measured in meters.

v_{o} - Initial velocity of the softball, measured in meters per second.

t - Time, measured in seconds.

g - Gravitational acceleration, measured in meters per square second.

If we know that y = y_{o}, t = 3.56\,s and g = -9.807\,\frac{m}{s^{2}}, the initial velocity of the softball is:

v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0

3\cdot v_{o} -44.132\,m= 0

v_{o} = 14.711\,\frac{m}{s}

The initial velocity of the softball is 14.711 meters per second.

8 0
3 years ago
A car moving west slows down from a velocity of 50 m/s to 20 m/s in 15 seconds. What is its acceleration?
velikii [3]
Vf = Vo + at

Vf = 20 m/s
Vo = 50 m/s
a = ? 
t = 15

Therefore

20 = 50 + 15a

20 - 50 = 15a

-30 = 15a

a = -30 / 15

a = -2 m/s²
6 0
3 years ago
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