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3 years ago

What effect does an unbalanced force have on an object?

Physics

2 answers:

Vladimir79 [104]3 years ago

7 0

Answer: This force will cause the motion in an object.

Explanation:

Unbalanced force is defined as the forces which are not equal in magnitude. When two forces with different magnitudes are applied on an object from opposite directions, the object will start moving in the direction of the net force.

These forces result in the motion of an object.

Balanced forces are the forces which are equal in magnitude. When two forces with equal magnitude are applied on an object from opposite directions, the object will not start moving.

These forces does not result in the motion of an object.

Hence, this force will cause the motion in an object.

Mars2501 [29]3 years ago

5 0

it either changes the direction of the object or it stops the object in motion.

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Answer:

vf = 0

Explanation:

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Therefore, his final velocity vf is

vf = 0

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What morbid structure traditionally has thirteen steps?

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3 years ago

Heat is added to a 2kg piece of ice at a rate of 793kW. How long will it take for ice to melt if it was initially 0?

Ede4ka [16]

Answer:

0.84 s

Explanation:

Step 1

Given information:

Mass of the ice (m) = 2.0 kg

Heat transfer rate (Q/T) = 793.0 kW

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$\frac{Q}{T} = \frac{mLf}{T}$

Substituting the corresponding values we have:

$793.0 kW= \frac{2.0kg(334 kJ/kg)}{T} \\ T = \frac{2.0kg(334 kJ/kg)}{793.0kW} = \frac{668kJ}{793kW} \\ = 0.84s$

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3 years ago

A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.

gtnhenbr [62]

Answer:

(a) vmax = 0.34m/s

(b) v = 0.13m/s

(d) x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

$v_{max}=\omega A$ (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

$\omega=\sqrt{\frac{k}{m}}$ (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

$v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}$

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

$v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}$

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

$x=Acos(\omega t)$

You solve the previous equation for t:

$t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s$

With this value of t, you can calculate the speed of the object with the following formula:

$v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}$

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

$\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s$

The position will be:

$x=Acos(\omega t)=0.04mcos((8.67rad/s)(0.021s))=0.039m$

The position of the object on which its speed is one-half its maximum velocity is 0.039

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4 years ago

Q1. Helmut rides 5km in 2h on his bike. His speed was *

baherus [9]

Answer:

Speed of bike = 2.5 km/h

Distance travel = 1,000 km (Approx.)

Explanation:

Given:

Distance cover by Helmut = 5 km

Time taken = 2 hour

Find:

Speed of bike

Computation:

Speed = Distance / Time

Speed of bike = 5 / 2

Speed of bike = 2.5 km/h

Given:

Speed of plane = 250 km/h

time taken = 3 hr 58 min = 3.967 hr

Find:

Distance travel

Computation:

Distance = Speed x time

Distance travel = 250 x 3.967

Distance travel = 991.669

Distance travel = 1,000 km (Approx.)

8 0

3 years ago