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jolli1 [7]
3 years ago
13

A 10.-kilogram object, starting at rest, slides down a frictionless incline with a constant acceleration of 2.0 ms for 40 second

s
What is the velocity of the object at the end of the 4.0 seconds?
Select one
a. 16 m/s
b. 20 mis
c. 8.0 m/s
Od 4.0 m/s
Check

Physics
1 answer:
Mashutka [201]3 years ago
6 0

The velocity of the object after 4.0 seconds is 8.0 m/s

Explanation:

The motion of the object is a uniformly accelerated motion (constant acceleration), therefore we can use the following  suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

For the object in this problem, we have:

u = 0 (it starts from rest)

a=2.0m/s^2 is the acceleration

t = 4.0 s is the time elapsed

Solving for v, we find the final velocity:

v=0+(2.0)(4.0)=8.0 m/s

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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The driver of a car travels at 90 km / h, observes some children playing on the road 50 m away, and applies the brakes, managing
Orlov [11]

Answer:

13,750 N

Yes

Explanation:

Given:

v₀ = 90 km/h = 25 m/s

v = 0 m/s

t = 4 s

Find: a and Δx

a = Δv / Δt

a = (0 m/s − 25 m/s) / (4 s)

a = -6.25 m/s²

F = ma

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6 0
3 years ago
1. The current in a wire is 0.72A. Calculate the charge that passes through the wire in; a. 4 s b. 60 s c. 180 s d. 7 s e. 0.5 s
Vlad [161]
Let current be I, charge be Q and time be t.
Here we are provided with,
I = 0.72A
t = 4s / 60s / 180s / 7s / 0.5s
We know,
I = Q/t

Case I
---------
When, t = 4s
0.72 = Q/4
Q = 0.72 * 4 = 2.88C

Case II
----------
When, t = 60s
0.72 = Q/60
Q = 0.72 * 60 = 43.2C

Case III
-----------
When, t = 180s
0.72 = Q/180
Q = 0.72 * 180 = 129.6C

Case IV
-----------
When, t = 7s
0.72 = Q/7
Q = 0.72 * 7 = 5.04C

Case V
----------
When, t = 0.5s
0.72 = Q/0.5
Q = 0.72 * 0.5 = 0.36C
6 0
3 years ago
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