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3 years ago

Velocity question...is it A B Cor D

2 answers:

6 0

In a constant acceleration of 3m per second, after 10 seconds,

3 x 10 = 30

B. 30m/s is your answer

hope this helps :D

Greeley [361]3 years ago

4 0

B is the answer
Hope it helped!

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A 1.50 3 103 - kg car starts from rest and accelerates uniformly to 18.0 m/s in 12.0 s. Assume that air resistance remains const

nexus9112 [7]

Answer:

Explanation:

let force exerted by engine be F.Net force =( F-400)N, applying newton law

F-400 = 1.5 x 10³x18 =27000 ,

F = 27400 N.

velocity after 12 s = 0 + 18 x 12 = 216 m/s

Average velocity = (0 + 216 )/2 = 108 m/s

Average power = force x average velocity = 27400 x 108 = 29.6 10⁵ W .⁶

b) At 12 s , velocity = 216 m/s

Instantaneous power = velocity x force = 216 x 27400 = 59.2 x 10⁶ W.

8 0

3 years ago

Read 2 more answers

As shown above an adhesive has been applied to contacting faces of two blocks so that the blocks interact with an adhesive force

rusak2 [61]

The magnitude of the adhesive force allows the top block to remain

attached to the bottom when the blocks and the wire are balanced.

The option that must be true is; $\mathbf{W_{bottom} = F_{adhesion}}$

Reason:

The tension exerted by the wire attached to the top block = T

Magnitude of the adhesive = $F_{adhesion}$

Weight of the top block = $W_{top}$

Weight of the bottom block = $W_{bottom}$

Given that with the exertion of the tension, the two blocks remain at rest, we have;

T = $W_{top}$ + $W_{bottom}$

The adhesive causes the bottom block to remain attached to the top block, we have;

Therefore, the magnitude of the adhesive force adds the bottom weight, to the top, weight, which gives;

The magnitude of the adhesive force = The weight of the bottom block

Therefore;

$W_{bottom}$ = $F_{adhesion}$

Learn more here:

brainly.com/question/18907970

8 0

3 years ago

Which test was conducted on the skin found under the victim’s nails?

BARSIC [14]

Y-STR analysis. It is a DNA test used to compute and match with the possibility via lap.

5 0

3 years ago

Read 2 more answers

A 45.0-N force pushes a cart 12.5 meters down a hallway. What is the work done on the cart?

makvit [3.9K]

Answer:

562.5J

Explanation:

The following were obtained from the question:

F = 45N

d = 12.5m

w =?

The work done can be achieved by using

w = F x d

w = 45 x 12.5

w = 562.5J

3 0

3 years ago

A ball is thrown straight up at time t=0 with an initial speed of 19m/s. Take the point of release to be y0=0 and upwards to be

Wittaler [7]

First we write the corresponding kinematics equations:
a = -g
v = -g * t + vo
y = -g * ((t ^ 2) / 2) + vo * t + yo
Substituting the values:
y = - (9.81) * (((0.50) ^ 2) / 2) + (19) * (0.50) + (0) = 8.27m
answer:
the displacement at the time of 0.50s is 8.27m

4 0

3 years ago