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Anna71 [15]
3 years ago
15

I can't figure out what the third number I need is in this question a car traveling at +20 m/s applies the brakes and stops in 6

.5s. what is the cars acceleration.
Physics
1 answer:
dlinn [17]3 years ago
7 0

The acceleration is -3.1 m/s^2

Explanation:

The acceleration of an object is the rate of change in velocity of the object.

Mathematically, it is given by

a=\frac{v-u}{t}

where

u is the initial velocity

v is the final velocity

t is the time taken for the velocity to change from u to v

For the car in this problem, we have:

u = +20 m/s

v = 0 (because the car comes to a stop)

t = 6.5 s (time taken)

Therefore, the acceleration is

a=\frac{0-(+20)}{6.5}=-3.1 m/s^2

Where the negative sign means that the direction of the acceleration is opposite to the direction of the initial velocity.

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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Keep in mind, the energy is conserved in a pendulum.
Here’s more information:
https://blogs.bu.edu/ggarber/interlace/pendulum/energy-in-a-pendulum/
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3 years ago
PLZ HELPPPPP!! i'll give brainliest
kati45 [8]

Answer:

12N to the right.

Explanation:

There is a force of 12N upwards and a force of 12N downwards: these cancel out.

Assign a negative value to forces towards the left, and a positive value to the forces towards the right: -3N and +15N

Combine them: -3N+15N = 12N

The net force has a magnitude of 12N, and since our answer was positive, it acts towards the right.

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3 years ago
A wooden block with mass 1.60 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 30.0° (p
andreyandreev [35.5K]

Answer:

The amount of potential energy that was initially stored in the spring is 88.8 J.

Explanation:

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Mass of block = 1.60 kg

Angle = 30.0°

Distance = 6.55 m

Speed = 7.50 m/s

Coefficient of kinetic friction = 0.50

We need to calculate the amount of potential energy

Using formula of conservation of energy between point A and B

U_{A}+k_{A}+w_{A}=U_{B}+k_{B}

U_{A}+0-fd=mgy+\dfrac{1}{2}mv^2

U_{A}=\mu mg\cos\theta\times d+mg h\sin\theta+\dfrac{1}{2}mv^2

Put the value into the formula

U_{A}=0.50\times1.60\times9.8\cos30\times6.55+1.60\times9.8\times6.55\sin30+\dfrac{1}{2}\times1.60\times(7.50)^2

U_{A}=88.8\ J

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3 years ago
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3 years ago
a load of 800 newton is lifted by an effort of 200 Newton. if the load is placed at a distance of 10 cm from the fulcrum. what w
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Answer:

40 cm

Explanation:

We are given that

Load=800 N

Effort=200 N

Load  distance=10 cm

We have to find the effort distance.

We know that

load\times load\;distance=Effort\times effort\;distance

Using the formula

800\times 10=200\times effort\;distance

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Effort distance=\frac{8000}{200}

Effort distance=40 cm

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