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3 years ago

5. A ball is sitting at rest on the floor. What will happen if a balanced force is applied to the ball? *

Physics

1 answer:

Leno4ka [110]3 years ago

3 0

When the balanced force is applied on the ball It will roll away from the force.

<u>Explanation:</u>

A ball lies on the floor in rest. If the balanced force is applied to

the ball, the force will push away.

The forces would include gravity and the forces of air particles entering the ball from almost all directions.
And the ground is exercising the force and shifting away from the impact.

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Suppose your local apparent solar time is 11 am, and you have a clock that says it is noon in Greenwich. What is your longitude

jeyben [28]

The longitude based on the time difference is 15 degrees.

<h3>Longitude of complete rotation of the Earth</h3>

The longitude of a complete rotation of the earth in a 24 hours is calculated as follows;

$= \frac{360^0}{24} = 15^0 \ of \ longitude$

<h3>Time difference</h3>

The time difference between the local apparent solar time and the Greenwich time is calculated as follows;

$t = 12 pm \ - \ 11 am\\\\t = 1 \ hour$

Since it is one hour time difference, the longitude is 15 degrees.

Learn more about Earth longitude here: brainly.com/question/1939015

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2 years ago

Which electromagnetic wave has the lowest frequencies (less than 3×109 hertz)?

ICE Princess25 [194]

Answer:

Radio waves

Explanation:

The electromagnetic spectrum includes all different types of waves, which are usually classified depending on their frequency. Ordering them from the highest frequency to the lowest frequency, they are:

- Gamma rays

- X-rays

- Ultraviolet

- Visible light

- Infrared radiation

- Microwaves

- Radio waves

Radio waves are the electromagnetic waves with lowest frequency, their frequency is lower than 300 GHz ( $3\cdot 10^{11} Hz$ ) and therefore they are the electromagnetic waves with lowest energy (in fact, the energy of an electromagnetic wave is proportional to its frequency). They are generally used for radio and telecommunications since this type of waves can travel up to long distances.

5 0

4 years ago

Read 2 more answers

A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.7 1010 m (inside the orbit

Lubov Fominskaja [6]

Answer:

58515.9 m/s

Explanation:

We are given that

$d_1=4.7\times 10^{10} m$

$v_i=9.5\times 10^4 m/s$

$d_2=6\times 10^{12} m$

We have to find the speed (vf).

Work done by surrounding particles=W=0 Therefore, initial energy is equal to final energy.

$K_i+U_i=K_f+U_f$

$\frac{1}{2}mv^2_i-\frac{GmM}{d_1}=\frac{1}{2}mv^2_f-\frac{GmM}{d_2}$

$\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2}=\frac{1}{2}v^2_f$

$v^2_f=2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})$

$v_f=\sqrt{2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})}$

Using the formula

$v_f=\sqrt{v^2_i+2GM(\frac{1}{d_2}-\frac{1}{d_1})}$

$v_f=\sqrt{(9.5\times 10^4)^2+2\times 6.7\times 10^{-11}\times 1.98\times 10^{30}(\frac{1}{6\times 10^{12}}-\frac{1}{4.7\times 10^{10})}$

Where mass of sun= $M=1.98\times 10^{30} kg$

$G=6.7\times 10^{-11}$

$v_f=58515.9 m/s$

4 0

3 years ago

Need help 8th grade work

Alisiya [41]

1) Ecology

2) Food Web

3) Trophic Level

4) Producer

5) Autotroph

6) Consumer

7) Heterotroph

8) Decomposer

Hope tHis Helps ._.

6 0

3 years ago

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A particle with an initial linear momentum of 2.00 kg-m/s directed along the positive x-axis collides with a second particle, wh

ladessa [460]

Answer:

a) p₂ = 1.88 kg*m/s

θ = 273.4 º

b) Kf = 37% of Ko

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved.
Since momentum is a vector, their components (projected along two axes perpendicular each other, x- and y- in this case) must be conserved too.
The initial momenta of both particles are directed one along the x-axis, and the other one along the y-axis.
So for the particle moving along the positive x-axis, we can write the following equations for its initial momentum:

$p_{o1x} = 2.00 kg*m/s (1)$

$p_{o1y} = 0 (2)$

We can do the same for the particle moving along the positive y-axis:

$p_{o2x} = 0 (3)$

$p_{o2y} = 4.00 kg*m/s (4)$

Now, we know the value of magnitude of the final momentum p1, and the angle that makes with the positive x-axis.
Applying the definition of cosine and sine of an angle, we can find the x- and y- components of the final momentum of the first particle, as follows:

$p_{f1x} = 3.00 kg*m/s * cos 45 = 2.12 kg*m/s (5)$

$p_{f1y} = 3.00 kg*m/s sin 45 = 2.12 kg*m/s (6)$

Now, the total initial momentum, along these directions, must be equal to the total final momentum.
We can write the equation for the x- axis as follows:

$p_{o1x} + p_{o2x} = p_{f1x} + p_{f2x} (7)$

We know from (3) that p₀₂ₓ = 0, and we have the values of p₀1ₓ from (1) and pf₁ₓ from (5) so we can solve (7) for pf₂ₓ, as follows:

$p_{f2x} = p_{o1x} - p_{f1x} = 2.00kg*m*/s - 2.12 kg*m/s = -0.12 kg*m/s (8)$

Now, we can repeat exactly the same process for the y- axis, as follows:

$p_{o1y} + p_{o2y} = p_{f1y} + p_{f2y} (9)$

We know from (2) that p₀1y = 0, and we have the values of p₀₂y from (4) and pf₁y from (6) so we can solve (9) for pf₂y, as follows:

$p_{f2y} = p_{o1y} - p_{f1y} = 4.00kg*m*/s - 2.12 kg*m/s = 1.88 kg*m/s (10)$

Since we have the x- and y- components of the final momentum of the second particle, we can find its magnitude applying the Pythagorean Theorem, as follows:

$p_{f2} = \sqrt{p_{f2x} ^{2} + p_{f2y} ^{2} } = \sqrt{(-0.12m/s)^{2} +(1.88m/s)^{2}} = 1.88 kg*m/s (11)$

We can find the angle that this vector makes with the positive x- axis, applying the definition of tangent of an angle, as follows:

$tg \theta = \frac{p_{2fy} }{p_{2fx} } = \frac{1.88m/s}{(-0.12m/s} = -15.7 (12)$

The angle that we are looking for is just the arc tg of (12) which measured in a counter-clockwise direction from the positive x- axis, is just 273.4º.

Assuming that both masses are equal each other, we find that the momenta are proportional to the speeds, so we find that the relationship from the final kinetic energy and the initial one can be expressed as follows:

$\frac{K_{f}}{K_{o} } = \frac{v_{f1}^{2} + v_{f2} ^{2}}{v_{o1}^{2} + v_{o2} ^{2} } = \frac{12.5}{20} = 0.63 (13)$

So, the final kinetic energy has lost a 37% of the initial one.

6 0

3 years ago