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3 years ago

A 1,200 kg subcompact car accelerates at a rate of 3.0m/s

1 answer:

6 0

Missing question (found on internet):
"what is the force of the car?"

Solution:
According to Newton's second law, the force of the car is equal to the product between its mass and its acceleration:
$F=ma$
For the car in the problem,
$m=1200 kg$
$a=3.0 m/s^2$
So the force that accelerates the car is
$F=(1200 kg)(3.0 m/s^2)=3600 N$

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Irina18 [472]

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4 years ago

The centripetal force of an object of mass m is given by F(r)=mv2r, where v is the speed of rotation and r is the distance from

mr Goodwill [35]

Answer:

$\frac{dF}{dr} = -4.82$

Explanation:

As we know that the centripetal force is given as

$F = \frac{mv^2}{r}$

now we will have to find the rate of change in force with respect to its radial distance

so we have

$\frac{dF}{dr} = - \frac{mv^2}{r^2}$

so we have

m = 1000 kg

v = 13.89 m/s

r = 200 m

now we have

$\frac{dF}{dr} = -\frac{1000(13.89^2)}{200^2}$

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3 0

3 years ago

What is the correct answer?

pentagon [3]

Answer:

B) x^2+6x+8

Explanation:

x-4 | x^3+2x^2-16x-32

- x^3-4x^2 <-- (x-4)(x^2)

_________________

6x^2-16x-32

- 6x^2-24x <-- (x-4)(6x)

_________________

8x-32

- 8x-32 <- (x-4)(8)

___________________________

0 | x^2+6x+8

This means the answer is B) x^2+6x+8

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3 years ago

Andrew has a four-year college loan for $20,000. The lender charges a simple interest rate of 5 percent. How much interest will

atroni [7]

Answer:

B. $4,000

Explanation:

The formula for the simple interest is

$interest = P\cdot r \cdot t$

where

P is the principal

r is the interest rate

t is the number of years

In this problem,

P = 20,000 $

r = 0.05

t = 4 years

So the interest is

$P=(20,000 $)(0.05 )(4)=4,000 $$

7 0

4 years ago

While i was jogging, i wad able to run 5 miles in thirty minutes. from this information, we can determine my?

Llana [10]

You can only determine the speed since the only info we know is how much you ran in how long of a time.

7 0

3 years ago

Read 2 more answers