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Scrat [10]
3 years ago
12

A drag racer starts her car from rest and

Physics
1 answer:
olga nikolaevna [1]3 years ago
6 0

Answer:

10.9 s

Explanation:

Given:

Δx = 367 m

v₀ = 0 m/s

a = 6.2 m/s²

Find: t

Δx = v₀ t + ½ at²

367 m = (0 m/s) t + ½ (6.2 m/s²) t²

t ≈ 10.9 s

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Displacement is a vector magnitude that depends on the position of the body which is individualistic for the trajectory.

While, Distance is a scalar magnitude that measures over the trajectory.

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How do creepers explode
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From you getting close to them

Explanation:

Because its big brain time.

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The equation Bold r (t )equals(8 t plus 9 )Bold i plus (2 t squared minus 8 )Bold j plus (6 t )Bold k is the position of a parti
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Explanation:

It is given that, the position of a particle as as function of time t is given by :

r(t)=(8t+9)i+(2t^2-8)j+6tk

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v=\dfrac{d[(8t+9)i+(2t^2-8)j+6tk]}{dt}

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a=\dfrac{dv(t)}{dt}

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a=(4j)\ m/s^2

At t = 0, v=(8i+0+6k)\ m/s

v(t)=\sqrt{8^2+6^2} =10\ m/s

Hence, this is the required solution.

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3 years ago
Astronomical observatories have been available since ancient times, and many cultures set aside special sites for astronomical o
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Telescope

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3 years ago
A factory worker pushes a crate of mass 31.0 kg a distance of 4.35 m along a level floor at constant velocity by pushing horizon
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Answer:

a. 79.1 N

b. 344 J

c. 344 J

d. 0 J

e. 0 J

Explanation:

a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force F_p by the worker must be equal to the friction force F_f on the crate, which is the product of friction coefficient μ and normal force N:

Let g = 9.81 m/s2

F_p = F_f = \mu N = \mu mg = 0.26 * 31 * 9.81 = 79.1 N

b. The work is done on the crate by this force is the product of its force F_p and the distance traveled s = 4.35

W_p = F_ps = 79.1*4.35 = 344 J

c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

W_f = F_fs = -79.1*4.35 = -344 J

This work is negative because the friction vector is in the opposite direction with the distance vector

d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.

e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction

W_p + W_f = 344 - 344 = 0 J

8 0
3 years ago
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