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3 years ago

How fast is a car going in m/5 if speedometer reads 45 miles?

Physics

1 answer:

fredd [130]3 years ago

4 0

Answer:

i think you would have to divide to 2 numbers to get the sum

Explanation:

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A ball is thrown horizontally from the top of a 55 m building and lands 150 m from the base of the building. Ignore air resistan

PtichkaEL [24]

Answer:

a) t =3.349 s

b) V_x,i = 44.8 m/s

c) V_y,f = 32.85 m/s

d) V = 55.55 m/s

Explanation:

Given:

- Total throw in x direction x(f) = 150 m

- Total distance traveled down y(f) = 55 m

Find:

a) How long is the rock in the air in seconds.

b) What must have been the initial horizontal component of the velocity, in meters per second?

c) What is the vertical component of the velocity just before the rock hits the ground, in meters per second?

d) What is the magnitude of the velocity of the rock just before it hits the ground, in meters per second?

Solution:

- Use the second equation of motion in y direction:

y(f) = y(0) + V_y,i*t + 0.5*g*t^2

- V_y,i = 0 (horizontal throw)

55 = 0 + 0 + 0.5*(9.81)*t^2

t = sqrt ( 55 * 2 / 9.81 )

t =3.349 s

- Use the second equation of motion in x direction:

x(f) = x(0) + V_x,i*t

150 = 0 + V_x,i*3.349

V_x,i = 150 / 3.349 = 44.8 m/s

- Use the first equation of motion in y direction:

V_y,f = V_y,i + g*t

V_y,f = 0 + 9.81*3.349

V_y,f = 32.85 m/s

- The magnitude of velocity of ball when it hits the ground is:

V^2 = V_y,f^2 + V_x,i^2

V = sqrt (32.85^2 + 44.8^2)

V = 55.55 m/s

5 0

3 years ago

In a 400-m relay race the anchorman (the person who runs the last 100 m) for team A can run 100 m in 9.8 s. His rival, the ancho

melisa1 [442]

Answer:

largest lead = 3 m

Explanation:

Basically, this problem is about what is the largest possible distance anchorman for team B can have over the anchorman for team A when the final leg started that anchorman for team A won the race. This show that anchorman for team A must have higher velocity than anchorman for team B to won the race as at the starting of final leg team B runner leads the team A runner.

So, first we need to calculate the velocities of both the anchorman

given data:

Distance = d = 100 m

Time arrival for A = 9.8 s

Time arrival for B = 10.1 s

Velocity of anchorman A = D / Time arrival for A

=100/ 9.8 = 10.2 m/s

Velocity of anchorman B = D / Time arrival for B

=100/10.1 = 9.9 m/s

As speed of anchorman A is greater than anchorman B. So, anchorman A complete the race first than anchorman B. So, anchorman B covered lower distance than anchorman A. So to calculate the covered distance during time 9.8 s for B runner, we use

d = vt

= 9.9 x 9.8 = 97 m

So, during the same time interval, anchorman A covered 100 m distance which is greater than anchorman B distance which is 97 m.

largest lead = 100 - 97 = 3 m

So if his lead no more than 3 m anchorman A win the race.

5 0

3 years ago

if a man hits a golf ball (0.2 kg) which accelerates at a rate of 20 m/s squared. what amount of force acted on the ball?

kkurt [141]

F = ma

We have mass = 0.2kg

and acceleration = 20 m/s^2

So..

F = (0.2)(20)

F = 4 N

6 0

3 years ago

A transformer supplies 60 watts of power to a device that is rated at 20 volts. The primary coil is connected to a 120-volt ac s

Nataly [62]

Answer:

I = 0.5 A

Explanation:

Given: P=60 Watts, Voltage supply V = 120 Volts (for primary coil)

Solution:

we have P = V I

⇒ I = P /V = 60 Watts / 120 Volts

I = 0.5 A

4 0

3 years ago

When the potential difference between the plates of a capacitor is increased by 3.50 V , the magnitude of the charge on each pla

hodyreva [135]

Answer:

42.9 μF

Explanation:

V = 3.50 V, Q = 150 μC

C = Q/V = 150/3.50 μF = 42.9 μF

5 0

3 years ago