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BlackZzzverrR [31]
2 years ago
14

Which force stops the car from moving?

Physics
2 answers:
Alex73 [517]2 years ago
8 0

Answer:

The force of friction.

Explanation:

Gravity keeps the car on the ground.

Motion Allows the car to move.

The force of speed doesnt make sense.

Friction would cause the car to stop moving.

yawa3891 [41]2 years ago
3 0
Yes, friction is the force enabling the car to move along the road.
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A bird flies 3.6 km due west and then 1.8 km due north. Another bird flies 1.8 km due west and 3.6 km due north. What is the ang
kondor19780726 [428]
Define unit vectors as follows:
\hat{i} is in the eastern direction.
\hat{j} is in the northern direction.

The position of the first bird is
\vec{a} = -3.6 \, \hat{i} + 1.8 \, \hat{j}

The position of the second bird is
\vec{b} = - 1.8 \, \hat{i} + 3.6 \, \hat{j}

Let θ = the angle between the net displacement vector for the two birds.
By definition,
\vec{a} . \vec{b} = |a| |b| cos\theta \\\\ \theta = cos^{-1} ( \frac{\vec{a}.\vec{b}}{|a||b|} )

\vec{a}.\vec{b} = (-3.6)(-1.8)+(3.6)(1.8) = 12.96
|a| =  \sqrt{3.24+12.96} =4.025
Similarly,
|b| = 4.025

Therefore
\theta = cos^{-1}  \frac{12.96}{4.025^{2}} =36.9^{o}

Answer:  36.9°
5 0
3 years ago
A person on a merry-go-round makes a complete revolution in about 3.17 s.
GREYUIT [131]

Answer:

second one

Explanation:

HOPE THIS helps

8 0
2 years ago
How large a force is necessary to stretch a 4.0-mm-diameter steel wire from its original length by 1.0%?
jekas [21]

The force needed to stretch the steel wire by 1% is 25,140 N.

The given parameters include;

  • diameter of the steel, d = 4 mm
  • the radius of the wire, r = 2mm = 0.002 m
  • original length of the wire, L₁
  • final length of the wire, L₂ = 1.01 x L₁ (increase of 1% = 101%)
  • extension of the wire e = L₂ - L₁ = 1.01L₁ - L₁ = 0.01L₁
  • the Youngs modulus of steel, E = 200 Gpa

The area of the steel wire is calculated as follows;

A = \pi r^2\\\\ A= 3.142 \times (0.002)^2\\\\ A= 1.257 \times 10^{-5} \ m^2

The force needed to stretch the wire is calculated from Youngs modulus of elasticity given as;

E = \frac{stress}{strain} = \frac{F/A}{e/L} = \frac{FL}{Ae} \\\\F = \frac{EAe}{L}

F = \frac{200 \times 10^9\  \times\  1.257\times 10^{-5}\  \times \ 0.01l_1}{l_1} \\\\F = 25,140\ N

Thus, the force needed to stretch the steel wire by 1% is 25,140 N.

Learn more here: brainly.com/question/21413915

4 0
2 years ago
Amanda spent 2/5 of her time after school doing homework and ¼ of her remaining time riding her bike. If she rode her bike for 4
Doss [256]

Answer:120 min

Explanation:

Given

Amanda  spent \frac{2}{5} of her time after school doing Home work

And \frac{1}{4} of her remaining  time riding her bike

It is given that she rode her bike for 45 minutes in a week

Let t be the time after school

therefore Amanda spend \frac{2t}{5} in home work and  \frac{3t}{5} time is left

From remaining \frac{3t}{5} time she spends \frac{1}{4} time riding her bike

therefore \frac{3t}{5}\times \frac{1}{4}=45

thus t=300 min

therefore time  spent on home work is \frac{2}{5}\times 300=120 min

6 0
3 years ago
A space probe has two engines. Each generates the same amount of force when fired, and the directions of these forces can be in-
Leto [7]

Answer:

t = 39.60 s

Explanation:

Let's take a careful look at this interesting exercise.

In the first case the two motors apply the force in the same direction

            F = m a₀          

           a₀ = F / m

with this acceleration it takes t = 28s to travel a distance, starting from rest

           x = v₀ t + ½ a t²

           x = ½ a₀ t²

           t² = 2x / a₀

           28² = 2x /a₀          (1)

in a second case the two motors apply perpendicular forces

we can analyze this situation as two independent movements, one in each direction

           

in the direction of axis a, there is a motor so its force is F/2

               

the acceleration on this axis is

          a = F/2m

          a = a₀ / 2

so if we use the distance equation

             x = v₀ t + ½ a t²

as part of rest v₀ = 0

             x = ½ (a₀ / 2) t²

             

let's clear the time

             t² = (2x / a₀)  2

we substitute the let of equation 1

             t² = 28² 2

             t = 28 √2

             t = 39.60 s

4 0
3 years ago
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