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3 years ago

A car goes from point A to point B, five miles away and then returns to point A. The car is going 15 mph.

Physics

1 answer:

Alex73 [517]3 years ago

3 0

Answer: B) 0 mph

Explanation:

Velocity is a vector quantity. It can be measured as rate of change of displacement.

$velocity\frac{Displacement}{time}$

Displacement is the straight path length between final and initial position of the body in motion.

It is given that the car goes from point A to point B and the returns to Point A. Hence, the net displacement of the car is 0 as the car comes back to its initial position.

Therefore, the velocity of the car is 0 mph.

Option B is correct.

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Please help me solve all of them ( a, b, c and d ) thankiew !! <br> I’m also kind of in a rush

Sergio039 [100]

Answer:

V= IR

9V = I ×( 12+6)

I = 9/ 18 A = 0.5 A

V=IR

240 = 6 A ×( 20 + R)

40 = 20 + R

R = 20 ohm

resultant resistance of the 2 parallel resistances= Ro

1/Ro = 1/ 5 + 1/ 20

1/Ro =( 20+5)/100

= 1/Ro = 1/4

Ro= 4 ohm

V=IR

V = 2A × ( 1+ 4 OHM)

V = 10V

equivalent resistance = Ro

1/Ro = 1/(2+8) + 1/(5+5)

1/Ro = 1/10 +1/10

2/10 = 1/ Ro

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2 years ago

A positive charge of 8.0 × 10-4 C is in an electric field that exerts a force of 3.5 × 10-4 N on it. What is the strength of the

Gennadij [26K]

Answer:

E = 0.437 N/C

Explanation:

Given that,

Charge, $q=8\times 10^{-4}\ C$

Electric force, $F=3.5\times 10^{-4}\ N$

Let the strength of the electric field is E. We know that, the electric force is given by :

F = qE

Where

E is the electric field strength

$E=\dfrac{F}{q}\\\\E=\dfrac{3.5\times 10^{-4}}{8\times 10^{-4}}\\E=0.437\ N/C$

So, the strength of the electric field is equal to 0.437 N/C.

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3 years ago

Why don't we use more renewable energy sources

sattari [20]

Answer:

Renewable energy sources 1.Cost more then nonrenewable resources and 2.Are so far not as reliable as nonrenewable energy.

Explanation:

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3 years ago

Read 2 more answers

A police car is driving down the street with it's siren on. You are standing still on the sidewalk beside the street. If the fre

diamong [38]

Answer:

A) 1568.60 Hz

Explanation:

This change is frequency happens due to doppler effect

The Doppler effect is the change in frequency of a wave in relation to an observer who is moving relative to the wave source

$f_(observed)=\frac{(c+-V_r)}{(C+-V_s)} *f_(emmited)\\$

where

C = the propagation speed of waves in the medium;

Vr= is the speed of the receiver relative to the medium,(added to C, if the receiver is moving towards the source, subtracted if the receiver is moving away from the source;

Vs= the speed of the source relative to the medium, added to C, if the source is moving away from the receiver, subtracted if the source is moving towards the receiver.

A) Here the Source is moving towards the receiver(C-Vs)

and the receiver is standing still (Vr=0) therefore the observed frequency should get higher

$f_(observed)=\frac{C}{C-V_s} *f_(emmited)\\=\frac{343}{343-15}*1500\\ =1568.60 Hz$

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3 years ago

A Hooke's law spring is mounted horizontally over a frictionless surface. The spring is then compressed a distance d and is used

zloy xaker [14]

Answer:

The compression is $\sqrt{2} \ d$ .

Explanation:

A Hooke's law spring compressed has a potential energy

$E_{potential} = \frac{1}{2} k (\Delta x)^2$

where k is the spring constant and $\Delta x$ the distance to the equilibrium position.

A mass m moving at speed v has a kinetic energy

$E_{kinetic} = \frac{1}{2} m v^2$ .

So, in the first part of the problem, the spring is compressed a distance d, and then launch the mass at velocity $v_1$ . Knowing that the energy is constant.