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3 years ago

What happens to the direction of an object as an unbalanced force acts on it? Give an example

Physics

2 answers:

xz_007 [3.2K]3 years ago

8 0

Hello :)

Answer:

When an unbalanced force acts on an object, the side with the greater force's direction makes the object move along its direction. For example, when two cars collide on a head-on collision, both vehicles will move in the direction that the vehicle with the greater force is moving in.

Pie3 years ago

5 0

Answer:

When an unbalanced force acts on a body the side with the greater force's dircetion makes the object move along its direction

Also to find the net force acting on the bofldy you can subtract the two force acting on the body

In case of balanced force the net force will always be 0

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What is transferred by a force moving an object through a distance?

Komok [63]

<span>When you apply force to move an object at a distance, you are applying work. And work is energy in transit. The answer is letter D. For example, you see a cart at a distance. You observe that it is not moving. You want to transfer it to your backyard. You apply force to the cart and observed that the cart is not at the same position as it was before. You are applying work to the cart by transferring your energy to it.</span>

7 0

3 years ago

Read 2 more answers

10) Um viajante, ao desembarcar no aeroporto de Londres, observou que o valor da temperatura do ambiente na escala Fahrenheit é

joja [24]

Answer:

The observed temperature was 10º Celsius or 50º Fahrenheit.

Explanation:

The traveler observed that the temperature in Fahrenheit is five times the value of the temperature in Celsius, therefore:

$F = 5*C$

A Fahrenheit temperature relates to a Celsius one by the following expression:

$F = \frac{9}{5}*C + 32$

Using the second expression on the first, we can solve for the temperature in Celsius, this is done below:

$\frac{9}{5}*C + 32 = 5*C\\\frac{9}{5}*C - 5*C = -32\\\frac{9*C - 25*C}{5} = -32\\\frac{-16*C}{5} = -32\\-16*C = -32*5\\-16*C = -160\\C = \frac{-160}{-16} = 10\º\text{C}\\F = 5*C = 5*10 = 50\º\text{F}$

The observed temperature was 10º Celsius or 50º Fahrenheit.

7 0

3 years ago

At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m2. A fan is 30 m from the loudspeak

aliina [53]

Explanation:

Below is an attachment containing the solution.

5 0

3 years ago

As you stand by the side of the road, a car approaches you at a constant speed, sounding its horn, and you hear a frequency of 8

kap26 [50]

Answer: velocity of the car is 113.33m/s

Explanation:

From Doppler effect,

in the case which the source is moving towards the observer at rest

f2 = v/(v-vs) *f1

where f2 is the final observed frequency

f1 is the initial observed frequency

v = 340m/s (speed of sound in air)

vs = velocity of the source of sound.

rearranging the above equation

f2*(v - vs) = f1* v

vs = (f1* v/f2) - v

but f1 = 80Hz

f2 = 60Hz

v = 340m/s

substituting,

vs = (80 x 340)/60 - 340

vs = 453.33 - 340

vs = 113.33m/s

velocity of the car is 113.33m/s

5 0

4 years ago

A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci

ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity ( $v$ ), measured in meters per second, by using this kinematic equation:

$v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s}$ (1)

Where:

$a$ - Acceleration, measured in meters per square second.

$\Delta s$ - Travelled distance, measured in meters.

$v_{o}$ - Initial velocity, measured in meters per second.

If we know that $v_{o} = 0\,\frac{m}{s}$ , $a = 2.35\,\frac{m}{s^{2}}$ and $\Delta s = 595\,m$ , the final velocity of the rocket is:

$v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}$

$v\approx 52.882\,\frac{m}{s}$

The time associated with this launch ( $t$ ), measured in seconds, is:

$t = \frac{v-v_{o}}{a}$

$t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }$

$t = 22.503\,s$

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

$v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o})$ (2)

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

If we know that $v_{o} = 52.882\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ , $a = -9.807\,\frac{m}{s^{2}}$ and $s_{o} = 595\,m$ , then the maximum height reached by the rocket is:

$v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})$

$s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}$

$s = 737.577\,m$

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

$s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}$ (2)

Where:

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

If we know that $s_{o} = 595\,m$ , $v_{o} = 52.882\,\frac{m}{s}$ , $s = 0\,m$ and $a = -9.807\,\frac{m}{s^{2}}$ , then the time needed by the rocket is:

$0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}$

$-4.904\cdot t^{2}+52.882\cdot t +595 = 0$

Then, we solve this polynomial by Quadratic Formula:

$t_{1}\approx 17.655\,s$ , $t_{2} \approx -6.872\,s$

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

7 0

3 years ago