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3 years ago

A car traveling south is 200 kilometers from its starting point after 2 hours. What is the average velocity of the car?

Physics

2 answers:

MakcuM [25]3 years ago

6 0

The velocity of the car would be 100 kilometer per hour.

MrRissso [65]3 years ago

3 0

The answer is a. 100 kilometers /hour south

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6/10

g100num [7]

Answer:pounds

Explanation:

7 0

3 years ago

A 60 W light bulb is powered by a connection to a wall outlet with 120 V across the plug terminals. a) What is the current passi

Oxana [17]

Answer:

The current through the resistor is 0.5 A

Explanation:

Given;

power of the light bulb = 60 W

voltage in the wall outlet across the plug terminals = 120 V

power of the light bulb is the product of voltage in the wall outlet across the plug terminals and the current passing through the resistor.

power = voltage x current

$Current = \frac{power}{voltage} = \frac{60}{120} = 0.5 A$

Therefore, for a 60 W light bulb powered by a connection to a wall outlet with 120 V across the plug terminals, the current passing through the resistor is 0.5 A

7 0

3 years ago

Read 2 more answers

The velocity is zero but the speed is a nonzero quantity?

jenyasd209 [6]

The starting and ending points of the motion are the same.. . . . .

5 0

3 years ago

A particle with a charge of 3.00 elementary charges moves through a potential difference of 4.50 volts. What is the change in el

GuDViN [60]

Answer:

$7.2\cdot 10^{-19} J$

Explanation:

The change in electrical potential energy of a charged particle moving through a potential difference is given by

$\Delta U = q \Delta V$

where

q is the magnitude of the charge of the particle

$\Delta V$ is the potential difference

In this problem:

- the charge of the particle is 3.00 elementary charges, so

$q=3e=3\cdot 1.6\cdot 10^{-19} J=4.8\cdot 10^{-19}J$

- the potential difference is

$\Delta V=4.50 V$

So, the change in electrical potential energy is

$\Delta U=(1.6\cdot 10^{-19}C)(4.50 V)=7.2\cdot 10^{-19} J$

7 0

3 years ago

The froghopper Philaenus spumarius is supposedly the best jumper in the animal kingdom. To start a jump, this insect can acceler

Kay [80]

Answer:

Part a)

$v_f = 4 m/s$

Part b)

$t = 0.001 s$

Part c)

$d = 0.815 m$

Explanation:

Part a)

As we know that initially the grass hopper is at rest at the ground position

Now the acceleration is given as

$a = 4000 m/s^2$

distance of the legs that it stretched is given as

$s = 2.0 mm$

so we have

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2(4000)(0.002)$

$v_f = 4 m/s$

Part b)

time taken to reach this speed is given as

$v_f - v_i = at$

$4 - 0 = 4000 t$

$t = 0.001 s$

Part c)

as the grass hopper reach the maximum height its final speed would be zero

so we will have

$v_f^2 - v_i^2 = 2 a d$

$0 - 4^2 = 2(-9.81) d$

$d = 0.815 m$

5 0

3 years ago