All categories

4 years ago

Find the net force it would take to accelerate an 800 Kg car at a rate of 5 m/s2

Physics

1 answer:

icang [17]4 years ago

3 0

Answer:

3,500 N.

Explanation:

You might be interested in

Please help me with this one

katen-ka-za [31]

Both of the pictures show how one picture is I believe in fall and the other one is in the forest and maybe in spring or summer. Also, the two pictures show 2 different animals. And maybe hunting for 2 different foods for them to survive.

* Hopefully this helps:) Mark me the brainliest:)

5 0

3 years ago

Suppose a boat moves at 12.0m/s relative to the water. If the boat is in a river with the current directed east of 2.50m/s, what

inessss [21]

Answer:

(a) 14.5 m/s

(b) 9.5 m/s

Explanation:

Let the speed of the boat in the still water (i.e relative to ground) is $v_0$ .

Given that the speed of the boat relative to water, $v_r=12.0$ m/s.

Speed of the water current of the river relative to the ground, $u_0=2.50$ m/s towards east.

(a) When the boat is heading toward the east (in the same direction of the current of the river)

The relative velocity, $v_r$ , of the boat with respect to ground is

$v_r=v_0-u_0$

$\Rightarrow 12=v_0-2.5$

$\Rightarrow v_0=12+2.5=14.5$ m/s

Hence, the boat's speed relative to the ground when it is heading east, with the current, is 14.5 m/s.

(b) When the boat is heading toward the west (in the opposite direction of the current of the river)

The relative velocity, $v_r$ , of the boat with respect to ground is

$v_r=v_0-(-u_0)$

$\Rightarrow 12=v_0+2.5$

$\Rightarrow v_0=12-2.5=9.5$ m/s

Hence, the boat's speed relative to the ground when it is heading west, against the current, is 9.5 m/s.

8 0

4 years ago

Which of the following decrease when an object becomes colder?

il63 [147K]

I think the answer b

6 0

3 years ago

The x and y component of a force are 2 and -3N the force is

sasho [114]

The answer is c have a good day

4 0

4 years ago

7. A roller coaster’s velocity at the top of a hill is 10 m/s. Two seconds later it reaches the bottom ofthe hill with a velocit

zheka24 [161]

Answer:

7) $a=8m/s^2$

8) $a=-5m/s^2$

9) $a=-1.5m/s^2$

10) $v=29.4m/s$

Explanation:

For the problems 7, 8 and 9 we just apply the definition of acceleration, since no more information is given, which is:

$a=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{t_f-t_i}$

So for each problem we will have:

7) $a=\frac{26m/s-10m/s}{2s}=8m/s^2$

8) $a=\frac{10m/s-25m/s}{3s}=-5m/s^2$

9) $a=\frac{0m/s-15m/s}{10s}=-1.5m/s^2$

For the problem 10, we use the equation of velocity in accelerated motion:

$v=v_0+at$

Since the ball starts from rest and the acceleration is that of gravity (we take the downward direction positive), we have:

$v=(0m/s)+(9.8m/s)(3s)=29.4m/s$

5 0

3 years ago