All categories

2 years ago

Find the Acceleration of a car with the mass of 1200kg and a force of 11x10^3 N

Physics

1 answer:

larisa86 [58]2 years ago

7 0

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question we have

$a = \frac{11 \times {10}^{3} }{1200} \\ = 9.16666...$

We have the final answer as

Hope this helps you

You might be interested in

When copper combines with oxygen to form copper oxide, the charge of the copper ion is

Archy [21]

The copper and oxygen and oxygen atoms are not ions, and the bonds are more covalent than they are ionic.

8 0

3 years ago

Read 2 more answers

How do you change time to kilograms?

enot [183]

You can't change time to a measurement of distance

5 0

3 years ago

A switch can

oee [108]

Answer: a switch can do A, B and E

Explanation:

8 0

3 years ago

Behavioral Adaptations: Behavior that animals begin life with that helps them meet their needs is called ________.

Arturiano [62]

I believe the blank would simply be behaviour adaptations. Behavioural adaptations are behaviours that organisms demonstrate to help them better survive and reproduce in a habitat. Hope that helps!!

6 0

3 years ago

Read 2 more answers

It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass

vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

$v^{2}=u^{2}+2\cdot a \cdot s$

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/ $Sec^{2}$

s= Distance traveled before stop m

Case 1

u= 13 m/sec, v=0, s= 57.46 m, a=?

$0^{2} = 13^{2} + 2 \cdot a \cdot57.46$

a = -1.47 m/ $Sec^{2}$ (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/ $Sec^{2}$ (since same friction force is applied)

$v^{2} = 29^{2} - 2 \cdot 1.47 \cdot S$

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

4 0

2 years ago