All categories

2 years ago

What is the acceleration of a 1000 kg car subject to a 550 N net force

Physics

1 answer:

Xelga [282]2 years ago

4 0

Answer:

The acceleration of a 1000 kg car subject to a 550 N net force = 0.55 m/s^2

Explanation:

Given:

F = 550 N

m = 1000 kg

To Find:

a = ?

Solution:

So by the equation by Newton's 2nd Law of Motion,

F = m x a

550 N = 1000 kg x a

a = 550 N/ 1000 kg

a = 0.55 m/s^2

Therefore,

The acceleration of a 1000 kg car subject to a 550 N net force = 0.55 m/s^2

PLEASE MARK ME AS BRAINLIEST!!!

You might be interested in

Suppose you were asked to find the torque about point p due to the normal force n in terms of given quantities. which method of

igomit [66]

Answer:

T=Lnsin $\alpha$

Please check the attached

Explanation:

The torque can simply be calculated by multiplying the length of the rod by the perpendicular force n as shown in the attached figure.

Note that sin90=1

T=Lsin $\alpha$ (nsin90)

T=Lsin $\alpha$ xn

T=Lnsin $\alpha$

7 0

3 years ago

At a distance of 11 cm from a presumably isotropic, radioactive source, a pair of students measure 65 cps (cps = counts per seco

Alborosie

To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.

The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,

$A_1*m=M*A_2$

$A_i =Area$

M,m = Counts per second

Our radios are given by

$r_1 = 11cm$

$R_2 = 20cm$

$m = 65cps$

Therefore replacing we have that,

$A_1*m=M*A_2$

$4\pi r_1^2*m = M * 4\pi R_2^2 M$

$r^2*m=MR^2$

$M = \frac{m*r^2}{R^2}$

$M = \frac{65*11^2}{20^2}$

$M = 19.6625cps$

Therefore the number of counts expect at a distance of 20 cm is 19.66cps

7 0

3 years ago

PLEASE ANSWER FAST I WILL GIVE BRAINLIEST

TiliK225 [7]

Explanation:

The answer is:

A squirrel runs up the trunk of a tree.

3 0

3 years ago

Read 2 more answers

What is the KE if a 10kg mass traveling at 5 m/s

11111nata11111 [884]

Answer: KE = 25 J

Explanation: You must use the formula

KE = 1/2 m v²

to solve this problem.

KE = 1/2 (10 Kg) (5 m/s)

KE = 1/2 (50 kgm/s)

KE = 25 J

7 0

2 years ago

Describe a cell at the beginning of the Mitosis process.

ElenaW [278]

<span>Cells may appear inactive during this stage, but they are quite the opposite. This is the longest period of the complete cell cycle during which DNA replicates, the centrioles divide, and proteins are actively produced. </span>

4 0

2 years ago