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2 years ago

Which example best demonstrates how unbalanced forces change the speed of an object's motion?

1 answer:

4 0

I don’t see a picture but unbalanced forces could be two boys pushing with a combined force of 400 Newton’s but the surface of what the box is laying on being 600 meaning since the ground is creating a higher force in the form of friction it will slow the boys down. When forces are unbalanced it means that the object can not be still or moving at a constant speed when one force is greater by a significant amount the object either slows quickly or accelerates fast depending on which factor is greater.

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if a string of light goes out when one of the bulbs is removed, are the lights probably connected in a series circuit or a paral

Minchanka [31]

Yes it is because other wise the light would stay on

4 0

3 years ago

Read 2 more answers

Atmospheric pressure is due to the weight _______

Kryger [21]

Of the gravitational pull and other things like mass. Planet earth it,s self as you said sir.

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3 years ago

A wire 3.22 m long and 7.32 mm in diameter has a resistance of 11.9 mΩ. A potential difference of 33.7 V is applied between the

Scorpion4ik [409]

Answer:

(a) Current is 2831.93 A

(b) $8.40A/m^2$

Explanation:

Length of wire l = 3.22 m

Diameter of wire d = 7.32 mm = 0.00732 m

Cross sectional area of wire

$A=\pi r^2=3.14\times 0.00366^2=4.20\times 10^{-5}m^2$

Resistance $R=11.9mohm=11.9\times 10^{-3}ohm$

Potential difference V = 33.7 volt

(A) current is equal to

$i=\frac{V}{R}=\frac{33.7}{11.9\times 10^{-3}}=2831.93A$

(B) Current density is equal to

$J=\frac{i}{A}$

$J=\frac{2831.93}{4.20\times 10^{-5}}=8.40A/m^2$

$R=\frac{\rho l}{A}$

$11.9\times 10^{-3}=\frac{\rho \times 3.22}{4.20\times 10^{-5}}$

$\rho =15.52\times 10^{-9}ohm-m$

4 0

2 years ago

If the thrower takes 0.90 s to complete one revolution, starting from rest, what will be the speed of the discus at release?

Marysya12 [62]

Ω₀ = the initial angular velocity (from rest)
t = 0.9 s, time for a revolution
θ = 2π rad, the angular distance traveled

Let
α = the angular acceleration
ω = the final angular velocity

The angular rotation obeys the equation
(1/2)*(α rad/s²)*(0.9 s)² = (2π rad)
α = 15.514 rad/s²

The final angular velocity is
ω = (15.514 rad/s²)*(0.9 s) = 13.963 rad/s

If the thrower's arm is r meters long, the tangential velocity of release will be
v = 13.963r m/s

Answer: 13.963 rad/s

8 0

3 years ago

2. A ball is dropped from rest. The acceleration due to gravity is 10m/s? and the time it

d1i1m1o1n [39]

Answer:

STEP BY STEP

V = S/T

V= 10/5

V = 2 m/s

3 0

3 years ago