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Westkost [7]
3 years ago
15

Which velocity-time graph matches the position-time graph?

Physics
2 answers:
skelet666 [1.2K]3 years ago
7 0

The answer is Graph C. To explain, this is because as we look at the position vs time graph, we see that after the first second, it was 30 meters from the start. That would mean that it took 1 second to get to 30 meters. That is shown in Graph c

Leona [35]3 years ago
6 0

Explanation:

In this question, a position time graph is shown. We need to find the corresponding velocity time graph.

In position time graph, till 1 second the object is in uniform motion and reaches a distance of 30 meters. Then the object is at till 3 seconds. After 3 seconds, the object again reaches a distance of 40 meters uniformly.

Corresponding velocity-time graph is shown in figure (3). The object is moving initially with a speed of 30 m/s. After 1 second, it comes to rest till 3 seconds then it again moves. So, the correct option is (c). Hence, this is the required solution.

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What do electric and magnetic fields have in common?
Nastasia [14]

Electric field, an electric property associated with each point in space when charge is present in any form. The magnitude and direction of the electric field are expressed by the value of E, called electric field strength or electric field intensity or simply the electric field.

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7 0
3 years ago
A diode vacuum tube consists of a cathode and an anode spaced 5-mm apart. If 300 V are applied across the plates. What is the ve
MAXImum [283]

Answer:

Explanation:

There is electric field between the plates whose value is given by the following expression

electric field E =  V /d where V is potential between the plates and d is distance between them

E = 300 / 5 x 10⁻³

=  60 x 10³ N/c

Force on electron = q E where q is charge on the electron

F = 1.6 X 10⁻¹⁹ X 60 X 10³ = 96 X 10⁻¹⁶ N.

Acceleration a = force / mass

a = 96 x 10⁻¹⁶/ mass  = 96 x 10⁻¹⁶ / 9.1 x 10⁻³¹

= 10.55 x 10¹⁵ m / s²

For midway , distance travelled

s =  2.5 x 10⁻³ m

s      =  1\2 a t²

t = \sqrt{\frac{2s}{a\\ } }

= \sqrt{\frac{2\times2.5\times10^{-3}}{ 10.55\times10^{15}}

t = .474 x 10⁻¹⁸ s

For striking the plate time is calculated as follows

t = [tex]\sqrt{\frac{2\times5\times10^{-3}}{ 10.55\times10^{15}}[/tex]

t = 0.67 x 10⁻¹⁸ s

3 0
3 years ago
A 60.0-kg skier with an initial speed of 12.0 m/s coasts up a 2.50-mhigh rise and angle is 35 degrees.Find her final speed at th
Shalnov [3]
Wt. = Fg = m*g = 60kg * 9.8N/kg=588 N.= 
<span>Wt. of skier. </span>

<span>Fp=588*sin35 = 337 N.=Force parallel to </span>
<span>incline. </span>
<span>Fv = 588*cos35 = 482 N. = Force perpendicular to incline. </span>

<span>Fk = u*Fv = 0.08 * 482 = 38.5 N. = Force </span>
<span>of kinetic friction. </span>
<span>d =h/sinA = 2.5/sin35 = 4.36 m. </span>

<span>Ek + Ep = Ekmax - Fk*d </span>
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<span>Ek=0.5*60*12^2-588*2.5-38.5*4.36=2682 J. </span>
<span>Ek = 0.5m*V^2 = 2682 J. </span>
<span>30*V^2 = 2682 </span>
<span>V^2 = 89.4 </span>
<span>V = 9.5 m/s = Final velocity.</span>
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4 years ago
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Answer:

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