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ozzi
3 years ago
13

What is the magnitude of force required to accelerate a car of mass 1.7 * 10 ^ 3 kg by 4.75 m/s

Physics
1 answer:
Gnesinka [82]3 years ago
8 0

What is the magnitude of force required to accelerate a car of mass 1.7 × 10³ kg by 4.75 m/s²

Answer:

F = 8.075 N

Explanation:

Formula for force is;

F = ma

Where;

m is mass

a is acceleration

F = 1.7 × 10³ × 4.75

F = 8.075 N

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A boat takes off from a dock at 2.5 m/s and speeds up at 4.2 m/s squared for six seconds how far has the most traveled
GaryK [48]

The boat's position x relative to its starting point x_0=0 is determined by

x=x_0+v_0t+\dfrac12at^2

where v_0 is its initial velocity, a is its acceleration, and t is time. After t=6\,\mathrm s, the boat has traveled

x=\left(2.5\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(4.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2

\implies x=91\,\mathrm m

3 0
3 years ago
Help, It has multiple answers for this question.
evablogger [386]
The answer is ultra violet radiation. From the air
4 0
3 years ago
You are designing a generator with a maximum emf 8.0 V. If the generator coil has 200 turns and a cross-sectional area of 0.030
shutvik [7]

Answer:

7.1 Hz

Explanation:

In a generator, the maximum induced emf is given by

\epsilon= 2\pi NAB f

where

N is the number of turns in the coil

A is the area of the coil

B is the magnetic field strength

f is the frequency

In this problem, we have

N = 200

A=0.030 m^2

\epsilon=8.0 V

B = 0.030 T

So we can re-arrange the equation to find the frequency of the generator:

f=\frac{\epsilon}{2\pi NAB}=\frac{8.0 V}{2\pi (200)(0.030 m^2)(0.030 T)}=7.1 Hz

4 0
3 years ago
Read 2 more answers
A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o
gulaghasi [49]

A. 0.77 A

Using the relationship:

P=\frac{V^2}{R}

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega

The two light bulbs are connected in series, so their equivalent resistance is

R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega

The two light bulbs are connected to a voltage of

V  = 240 V

So we can find the current through the two bulbs by using Ohm's law:

I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A

B. 142.3 W

The power dissipated in the first bulb is given by:

P_1=I^2 R_1

where

I = 0.77 A is the current

R_1 = 240 \Omega is the resistance of the bulb

Substituting numbers, we get

P_1 = (0.77 A)^2 (240 \Omega)=142.3 W

C. 42.7 W

The power dissipated in the second bulb is given by:

P_2=I^2 R_2

where

I = 0.77 A is the current

R_2 = 72 \Omega is the resistance of the bulb

Substituting numbers, we get

P_2 = (0.77 A)^2 (72 \Omega)=42.7 W

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

P=\frac{E}{t}

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

7 0
3 years ago
Which substance is a combination of different atoms?
Lady_Fox [76]

Answer:

The answer is compound

Explanation:

heterogeneous mixture is wronggggg

5 0
3 years ago
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