Ask question

Ask question

All categories

4 years ago

5

if forces acting on an object are unbalanced, which factors may result from an unbalanced force? Check all that apply. The net f

orce is negative. The net force is zero. The net force is positive. There is a change in motion. There is no change in motion.

2 answers:

Alchen [17]4 years ago

8 0

The net force is negative, and there is a change in motion.

Nataly_w [17]4 years ago

8 0

<h3><u>Answer;</u></h3>

The net force is negative
The net force is positive
There is change in motion

<h3><u>Explanation;</u></h3>

<u>Balanced forces are forces that act in opposite directions and are equal in size. When an object is acted upon by unbalanced forces then the object stays still or at rest or continues moving at a constant velocity in the same direction.</u>
<u>Unbalanced forces cause an object or a body to change its motion or velocity and direction. Therefore, unbalanced force cause accelerations.</u>

<h3><u /></h3>

You might be interested in

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

A monkey running through the jungle goes 0.198 km straight to the East, then turns 15.8° deflection from straight East toward th

inn [45]

Answer:

$|\vec r|=339.82\ m$

$\theta=-6.67^o$

Explanation:

<u>Displacement </u>

It's a vector magnitude that measures the space traveled by a particle between an initial and a final position. The total displacement can be obtained by adding the vectors of each individual displacement. In the case of two displacements:

$\vec r=\vec r_1+\vec r_2$

Given a vector as its polar coordinates (r,\theta), the corresponding rectangular coordinates are computed with

$x=rcos\theta$

$y=rsin\theta$

And the vector is expressed as

$\vec z==$

The monkey first makes a displacement given by (0.198 km,0°). The angle is 0 because it goes to the East, the zero-reference for angles. Thus the first displacement is

$\vec r_1==\ km=\ m$

The second move is (145 m , -15.8°). The angle is negative because it points South of East. The second displacement is

$\vec r_2==\ m$

The total displacement is

$\vec r=\ m+\ m$

$\vec r=\ m$

In (magnitude,angle) form:

$|\vec r|=\sqrt{337.52^2+(-39.48)^2}=339.82\ m$

$\boxed{|\vec r|=339.82\ m}$

$\displaystyle tan\theta=\frac{-39.48}{337.52}=-0.1169$

$\boxed{\theta=-6.67^o}$

5 0

4 years ago

lbvjy [14]

Answer:

Impulse = 322.5[kg*m/s], the answer is D

Explanation:

This method it is based on the principle of momentum and the amount of movement; and used to solve problems involving strength, mass, speed and time.

If units of the SI are used, the magnitude of the impulse of a force is expressed in N * s. however, when remembering the definition of the newton.

$N*S=(kg*m/s^{2} )*s = kg*m/s$

Now replacing the values on the following equation that express the definition of impulse

$Impulse = Force * Time\\\\Impulse = 215 * 1.5 = 322.5 [kg*m/s]$

4 0

3 years ago

Read 2 more answers

PLEASE ANSWER FAST!! A scientist put red blood cells in water that contained salt. Over time, the red blood cells burst. What is

Nitella [24]

It is most likely true that there was a lower concentration of salt in the water than in the cells because when blood cells are put in a hypotonic solution such as pure water, the little to no salt concentration in the water causes the cells to swell and burst. This would occur because the water would try to dilute the solution inside of the blood cell and which would, therefore, cause it to burst. Hope this helps!

7 0

4 years ago

Read 2 more answers

If you begin with 40 grams of a radioactive isotope and end with 10 grams, how many half-lifes of the radioactive isotope have p

Rom4ik [11]

30 grams of radioactive isotope have passed.

8 0

3 years ago