All categories

3 years ago

A horizontal line on a velocity/time graph shows ____ acceleration.

Physics

2 answers:

Novay_Z [31]3 years ago

6 0

Answer:

Zero (0)

Explanation:

A horizontal line will be no movement of negative of positive acceleration

dusya [7]3 years ago

3 0

Answer:

Zero acceleration

Explanation:

In a velocity-time graph, the gradient of the line describes how the acceleration varies.When the gradient is positive, it means we have a line that is upward sloping thus the accelaration is positive.When we have a negative gradient, it means the line is downward sloping thus the accelaration is negative.When the gradient is zero, there is no slope, thus the accelaration is zero, the case of a horizontal line.

You might be interested in

Pls how to solve this problem. help would be appreciated

Bingel [31]

Answer:

80 m/s

Explanation:

Given:

a = -5 m/s²

v = 0 m/s

Δx = 640 m

Find: v₀

v² = v₀² + 2a(x − x₀)

(0 m/s)² = v₀² + 2(-5 m/s²) (640 m)

v₀ = 80 m/s

7 0

4 years ago

A 14.0 gauge copper wire of diameter 1.628 mm carries a current of 12.0 mA . A) What is the potential difference across a 1.80 m

Serga [27]

Answer: a) 139.4 μV; b) 129.6 μV

Explanation: In order to solve this problem we have to use the Ohm law given by:

V=R*I whre R= ρ *L/A where ρ;L and A are the resistivity, length and cross section of teh wire.

Then we have:

for cooper R=1.71 *10^-8* 1.8/(0.001628)^2= 11.61 * 10^-3Ω

and for silver R= 1.58 *10^-8* 1.8/(0.001628)^2=10.80 * 10^-3Ω

Finalle we calculate the potential difference (V) for both wires:

Vcooper=11.62* 10^-3* 12 * 10^-3=139.410^-6 V

V silver= 10.80 10^-3* 12 * 10^-3=129.6 10^-6 V

8 0

4 years ago

Read 2 more answers

A diver is on a board 1.80 m above

Virty [35]

Answer:

v = 6.95 m/s

Explanation:

Given that,

A diver is on a board 1.80 m above the water, s = 1.8 m

The initial speed of the diver, u = 3.62 m/s

Let v is the speed with which she hit the water. It will move under the action of gravity. Using the equation of motion as follows :

$v^2-u^2=2gs\\\\v=\sqrt{u^2+2gs} \\\\v=\sqrt{(3.62)^2+2(9.8)(1.8)} \\\\v=6.95\ m/s$

So, she will hit the water with a speed of 6.95 m/s.

8 0

3 years ago

Read 2 more answers

* CRIMINOLOGY*

Setler [38]

I think the answer is B.

Hope this helps.

5 0

3 years ago

Read 2 more answers

A car travelling at 5.0 m/s starts to speed up. After 3.0 s its velocity has increased to 11 m/s. a) What is its acceleration? (

lions [1.4K]

initially, the car is traveling at 5.0 m/s.

so, we know acceleration for changing velocity is :

a = (v-v $_{o}$ )/t ..........(i)

where v is the final velocity

v $_{o}$ is the initial velocity

t is the time taken to change velocity

Now, as per the question :

initial velocity, v $_{o}$ =5.0 m/s

final velocity, v =11 m/s

time taken, t = 3 s

putting the values in equation (i),

a = ( 11-5 )/3

a = 2 m/s²

Therefore, a, after 3 s, is <em>2 m/s².</em>

4 0

1 year ago