All categories

3 years ago

What is the acceleration of the object?______ m/s2

Physics

1 answer:

professor190 [17]3 years ago

8 0

Answer:

Explanation:

this graph show the decrease in velocity with time

at time t=0 velocity is 40 m/sec

at time t=12sec velocity is decreases from 40 to 10 m/s

so the acceleration of the object at time=12 sec is

a=v/t

a=Δv/t

Δv=v1-v2=40-10=30 m/s

a=30/12=2.5 m/s^2

You might be interested in

Question 1 of 10

Bad White [126]

The answer is b maybe?

3 0

3 years ago

A 26 kg body is moving through space in the positive direction of an x axis with a speed of 350 m/s when, due to an internal exp

Gnom [1K]

Answer:

a.) 1567.2 m/s

b.) 149.4 m/s

Explanation:

Given that a 26 kg body is moving through space in the positive direction of an x axis with a speed of 350 m/s when, due to an internal explosion, it breaks into three parts. One part, with a mass of 7.8 kg, moves away from the point of explosion with a speed of 180 m/s in the positive y direction. A second part, with a mass of 8.8 kg, moves in the negative x direction with a speed of 640 m/s.

The x-component of the third part can be calculated by assuming that it moves in a positive x axis.

The third mass = 26 - ( 7.8 + 8.8)

The third mass = 26 - 16.6

The third mass = 9.4kg

since momentum is conserved, the momentum before explosion will be equal to sum of the momentum after explosion

26 x 350 = -8.8 x 640 + 9.4V

9100 = -5632 + 9.4V

9.4V = 9100 + 5632

9.4V = 14732

V = 14732/9.4

V = 1567.2 m/s

(b) y-component of the velocity of the third part will be

7.8 x 180 = 9.4 V

1404 = 9.4V

V = 1404/9.4

V = 149.4 m/s

7 0

3 years ago

1. Calcula la fuerza de atracción electrostática entre dos cuerpos de cargas q1 = -18 C y q2 = +5 mC, separados entre sí por una

romanna [79]

Answer:

A) F=-20.16×10⁹N

B) if the distance doubles, force is 4 times smaller.

Explanation:

q1=-28C

q2=5mC=0.005C

d=25cm=0.25m

Electrostatic force between charges: F=k×q1×q2/d², where k is a coefficient that has the value k=9 × 10⁹ N⋅m²⋅C^(-2) for air.

Thus:

F=9×10⁹×(-28)×0.005/0.25²

F=-20.16×10⁹N

The minus sign indicates attraction.

If distance doubles, d1=2×d, then we have 4d² at the denominator and the force is 4 times smaller.

6 0

3 years ago

4. A car is accelerating at 40 m/s2, if the car is 400 kg how much force

Ksenya-84 [330]

Sorry I’m only in kindergarten is it 10kg must be supplied???

5 0

3 years ago

a body of a mass 1kg suspended from a spring is found to stretch the spring by 10cm. (A) what is the spring constant? what is pe

Nikitich [7]

Answer:

(A) The spring constant is <u>98 N/m.</u>

(B) The period of oscillation is <u>0.635 s.</u>

Explanation:

Given:

Mass of the body is, $m=1\ kg$

Extension length of the spring is, $x=10\ cm=0.1\ m$

Now, let 'k' be the spring constant.

The force acting on the body is due to gravity only and is equal to its weight.

So, weight of the body is given as:

$F_g=mg\\F_g=1\times 9.8\\F_g=9.8\ N$

Now, we know that for a spring-mass system, the net force acting on the body is equal to the product of the spring constant and extension length. Therefore,

$F_g=kx\\9.8=k(0.1)\\k=\frac{9.8}{0.1}=98\ N/m$