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3 years ago

Whenever a net force acts on an object, there is a change in the objects ____

Physics

1 answer:

maw [93]3 years ago

4 0

Velocity. Net force causes acceleration and acceleration causes a change in direction and/or magnitude of velocity

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A 5.2 cm diameter circular loop of wire is in a 1.35 T magnetic field. The loop is removed from the field in 0.29 sec. Assume th

Katyanochek1 [597]

Answer:

9.88 milivolt

Explanation:

Given: diameter d = 5.2 cm

magnetic field B_1 = 1.35 T, final magnetic field B_2 =0 T

t = 0.29 sec.

we know emf = - dΦ/dt

and flux Φ = BA

A= area

therefore emf ε = -A(B_2-B_1)/Δt

$=-\pi(d/2)^2\frac{B_2-B_1}{\Delta t} \\=-\pi(0.052/2)^2\frac{0-1.35}{0.29} \\=98.8\times10^4\\=9.88 mV$

3 0

3 years ago

A 50.0 kg shell is fired from a cannon on Earth's

ANTONII [103]

$ep = mgh$

$ep = (50.0)(9.81)(400)$

$ep = 196200 \: j$

I also calculated it with 9.8 as the gravity and the answer was 196000 J

5 0

3 years ago

In this circuit the battery provides 3 V, the resistance R1 is 7 Ω, and R2 is 5 Ω. What is the current through resistor R2? Give

sveta [45]

Answer:

The current pass the $R_2$ is $I = 0.25 A$

Explanation:

The diagram for this question is shown on the first uploaded image

From the question we are told that

The voltage is $V = 3V$

The first resistance is $R_1 = 7 \Omega$

The second resistance is $R_2 = 5 \Omega$

Since the resistors are connected in series their equivalent resistance is

$R_{eq} = R_1 +R_2$

Substituting values

$R_{eq} = 7 + 5$

$R_{eq} = 12 \Omega$

Since the resistance are connected in serie the current passing through the circuit is the same current passing through $R_2$ which is mathematically evaluated as

$I = \frac{V}{R_{eq}}$

Substituting values

$I = \frac{3}{12}$

$I = 0.25 A$

3 0

4 years ago

At what angle should the roadway on a curve with a 50m radius be banked to allow cars to negotiate the curve at 12 m/s even if t

kari74 [83]

Answer:

The road bank angle is 16.38⁰.

Explanation:

radius of curvature of the road, r = 50 m

allowable speed of car on the road, v = 12 m/s

The bank angle is calculated as;

$\theta = tan^{-1} (\frac{v^2}{gr} )$

where;

θ is the road bank angle

g is acceleration due to gravity = 9.8 m/s²

$\theta = tan^{-1} (\frac{v^2}{gr} )\\\\\theta = tan^{-1} (\frac{12^2}{9.8 \times 50} )\\\\\theta = tan^{-1} ( 0.2939)\\\\\theta = 16.38 ^0$

Therefore, the road bank angle is 16.38⁰.

8 0

3 years ago

A car changes speed from 25 m/s to 10 m/s in 240 seconds. Describe its acceleration.

Dmitry_Shevchenko [17]

Explanation:

Acceleration is change in velocity over change in time:

a = Δv / Δt

a = (10 m/s - 25 m/s) / (240 s - 0 s)

a = -0.0625 m/s²

So the car decelerates at 0.0625 m/s².

7 0

4 years ago