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DiKsa [7]
3 years ago
14

A force acts on a 5kg object at rest. How fast will the object accelerate on a frictionless surface?

Physics
1 answer:
MakcuM [25]3 years ago
8 0

Answer: The answer is C.) 25 m/s^2.

Explanation: If you input 5 as s, you would have to use the exponent 2.  This means that you have to multiply 5 by 5.  5 x 5= 25.  

Edit: Also, because the surface is frictionless, it will make the object go faster too.  Nothing can really slow it down unless something blocks it.

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A motorcyclist accelerates from rest to 10 mi/hr. what is the change in velocity
gulaghasi [49]

The change in velocity is 10 mi/h (4.47 m/s)

Explanation:

The change in velocity of the motorcyclist is given by

\Delta v = v-u

where

v is the final velocity

u is the initial velocity

In this problem, we have

u = 0 (the motorbike starts from rest)

v = 10 mi/h

Therefore, the change in velocity is

\Delta v = 10 -0 = 10 mi/h

And keeping in mind that

1 mile = 1609 m

1 h = 3600 s

We can convert it into m/s:

\Delta v = 10 \frac{mi}{h} \cdot \frac{1609 m/mi}{3600 s/h}=4.47 m/s

Learn more about velocity:

brainly.com/question/5248528

#LearnwithBrainly

6 0
3 years ago
Cool facts about satellites ???
svetoff [14.1K]
Satellites travel at 18,000 miles per hour. ...

A satellite gets better fuel economy than a Prius.

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7 0
2 years ago
Grade 11 question paper or memo for 2015
denis-greek [22]
What your saying doesn't make sense.
5 0
3 years ago
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He does whatever a spider can. Spider-Man, who has a mass of 76 [kg], is clinging onto an inclined wall forming an inclination a
Aliun [14]

Answer:570.54 N

Explanation:

Given

mass of man=76 kg

\theta =50^{\circ}

As man is standing over inclined building therefore

its weight has two components i.e. sin and cos component

Force perpendicular to inclined wall

F=mgcos\theta =76\times 9.8\times \sin 50

F=570.54 N

4 0
3 years ago
A point charge q1 is held stationary at the origin. A second charge q2 is placed at point a, and the electric potential energy o
likoan [24]

Answer:

U_{b}=+7.3*10^{-8}J

Explanation:

Given data

Electric potential at point a is Ua=5.4×10⁻⁸J

q₂ moves to point b where a negative work done on it  W_{a-b}=-1.9*10^{-8}J

Required

Electric potential energy Ub

Solution

When a particle moves from a point where the potential is Ua to a point where it is Ub the change in potential energy is equal to work done where the force exerted on the charge is conservative and work done is given by:

W_{a-b}=U_{a}-U_{b}\\U_{b}=U_{a}-W_{a-b}

Now  substitute the given values

So

U_{b}=5.4*10^{-8}J-(-1.9*10^{-8}J)\\U_{b}=+7.3*10^{-8}J

3 0
3 years ago
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