1. When an unbalanced force acts on an object,
O the object remains at rest
the weight of the object decreases.
O the object's motion changes.
the inertia of the object increases.

Answer:

bb kettle

Explanation:

it transfres electricsl to kinetic

Answer:

256 N

Explanation:

formula of centripetal force = mv²/r

m= 2kg

v= 8m/s

r= 0.5m

mv²/r = 2×8²/0.5 = 256N

The dependent variable is academic performance while the independent variable is presence/absence of tutorial support.

The correct results are:

In every experiment, there is a dependent and independent variable as well as an experimental and a control group.

The experimental group receive the treatment while the control group do not receive the treatment. The independent variable is manipulated and its impact on the dependent variable is evaluated.

The control group are students who did not receive the tutorial support while the experimental group are students that received the tutorial support.

The dependent variable in this case is academic performance. Its outcome depends on the presence or absence of tutorial support (independent variable).

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Answer:

1. 2.5×10¯⁹ N

2. 3.33×10¯¹¹ m/s²

Explanation:

1. Determination of the force of attraction.

Mass of astronaut (M₁) = 75 Kg

Mass of spacecraft (M₂) = 125000 Kg

Distance apart (r) = 500 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force of attraction (F) =?

The force of attraction between the astronaut and his spacecraft can be obtained as follow:

F = GM₁M₂ /r²

F = 6.67×10¯¹¹ × 75 × 125000 / 500²

F = 2.5×10¯⁹ N

Thus, the force of attraction between the astronaut and his spacecraft is 2.5×10¯⁹ N

2. Determination of the acceleration of the astronaut.

Mass of astronaut (m) = 75 Kg

Force (F) = 2.5×10¯⁹ N

Acceleration (a) of astronaut =?

The acceleration of the astronaut can be obtained as follow:

F = ma

2.5×10¯⁹ N = 75 × a

Divide both side by 75

a = 2.5×10¯⁹ / 75

a = 3.33×10¯¹¹ m/s²

Thus, the acceleration the astronaut is 3.33×10¯¹¹ m/s²

Answer:

B. kinetic energy

Explanation:

Thermal energy (It’s a low form of energy ) is a form of kinetic energy as it is produced as a result of motion of particles either if they vibrate at their position or they move along longer paths.

Answer: the answer would be C trust me i took the test if its not that its b

hope that helps

Explanation: i took the test

answer:

a) lower because the sound-wave fronts reach the platform at a frequency lower than the frequency at which they are produced

explanation :3

If the train is leaving the train station, then the people who are standing on the platform would hear a sound with a lower frequency since the train is moving further away. ^^

Answer:

a) 1.2 m/s

b) 0.350 m

c) 1.2 m/s

d) 0.250 m

Explanation:

a)

       [tex]v = \lambda * f (1)[/tex]

       [tex]v = \lambda * f = 6.0 m * 0.2 1/s = 1.2 m/s (3)[/tex]

b)

       ⇒ A = Δd/2 = 0.700m/2 = 0.350 m (4)

c)

d)

Answer:

c.

Equal to 200 N..........

Answer:

Explanation:

From the information given;

mass of the crate m = 41 kg

constant horizontal force = 135 N

where;

[tex]s_1 = 15.0 \ m \\ \\ s_2 = 12.0 \ m[/tex]

coefficient of kinetic friction [tex]u_k[/tex] = 0.28

a)

To start with the work done by the applied force [tex](W_f)[/tex]

[tex]W_F = F\times (s_1 +s_2) \times cos(0) \ J[/tex]

[tex]W_F = 135 \times (12 +15) \times cos(0) \ J \\ \\ W_F = (135 \times 37 )J \\ \\ W_F =4995 \ J[/tex]

Work done by friction:

[tex]W_{ff} = -\mu\_k\times m \times g \times s_2 \\ \\ W_{ff} = -0.320 \times 41 \times 9.81 \times 12 \ J \\ \\ W_{ff} = -1544.49 \ J[/tex]

Work done  by gravity:

[tex]W_g = mg \times (s_1+s_2) \times cos (90)} \ J \\ \\ W_g = 0 \ j[/tex]

Work done by normal force;

[tex]W_n = N \times (s_1 + s_2) \times cos (90) \ J[/tex]

[tex]W_n = 0 \ J[/tex]

b)

total work by all forces:

[tex]W = F \times (s_1 + s_2) + \mu_k \times m \times g \times s_2 \times 180 \\ \\ W = 135 \times (15+12) \ J - 0.320 \times 41 \times 9.81 \times 12[/tex]

W = 2100.5  J

c) By applying the work-energy theorem;

total work done = ΔK.E

[tex]W = \dfrac{1}{2}\times m \times (v^2 - u^2)[/tex]

[tex]2100.5 = 0.5 \times 41 \times v^2[/tex]

[tex]v^2 = \dfrac{2100.5}{ 0.5 \times 41 }[/tex]

[tex]v^2 = 102.46 \\ \\ v = \sqrt{102.46} \\ \\ \mathbf{v = 10.1 \ m/s}[/tex]

With rising heat, the steam pressure inside the pot builds up beyond atmospheric pressure, allowing the temperatures to rise well above boiling point. This design enables to save time, energy, and resources. The temperature inside a pressure cooker can well go beyond 110° C, which reduces the time needed to cook food.

Answer:

the bar is the top and bottem. the nucleas in the middle and the Spiral arm is the last space

Explanation:

Answer:

look pkch

Explanation:

?.............................

Answer:

The correct answer is A) It is upright and 1.5m behind the mirror

Explanation:

Your reflection must be upright, meaning vertical/erect, and the distance will be the exact same. Also, the reflected ray appears as if it had traveled from an object located behind the mirror.

Kinetic energy decreases if either the mass or the velocity of the object decreases or if both decrease. and  Kinetic energy increases if either the mass or the velocity of the object increases or if both increase.

Explanation:

These statements explain the relationship between mass, velocity, and kinetic energy .i.e KE = ½mv², (3) Kinetic energy decreases if either the mass or the velocity of the object decreases or if b decrease. (4)Kinetic energy increases if either the mass or the velocity of the object decreases or if both decrease.

When a body has a mass of m kg and travels a velocity of v m/s then the kinetic energy is given by,

KE=1/2mv²

In the above relation, it is clearly shown that,

Kinetic energy is directly proportional to the mass of the body,

i.e

KE∝m

and also Kinetic energy directly varies with the square of the velocity of the body.

I.e

KE∝v²

Therefore, From the given expression we can conclude that when mass and velocity increased either one of them or both, the kinetic energy also gets increased.

And when mass and velocity decreased either one of them or both, the kinetic energy also gets decreased. Hence options (3) and (4) are correct.

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Answer:

The torque exerted on the merry-go-round is 766.95 Nm

Explanation:

Given;

mass of the merry-go-round, m = 416 kg

radius of the disk, r = 1.7 m

angular speed of the merry-go-round, ω = 3.7 rad/s

time of motion, t = 2.9 s

The torque exerted on the merry-go-round is calculated as;

[tex]\tau = Fr= I\alpha\\\\\tau = (\frac{1}{2} m r^2)(\frac{\omega }{t} )\\\\\tau = (\frac{1}{2} \times 416 \times 1.7^2)( \frac{3.7}{2.9} )\\\\\tau = 766.95 \ Nm[/tex]

Therefore, the torque exerted on the merry-go-round is 766.95 Nm

Answer:

a place where two bones come together is known as a join

Answer:

Hey mate.....

Explanation:

This is ur answer....

Joints – A place in the body where bones come together. Non-Moveable Joints (sometimes called fixed or fibrous) – A place in the body where two or more bones come together but do not move.

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Answer:

B It is always increasing.

Explanation:

In Physics, entropy can be defined as the tendency or ability of a substance to reach maximum disorder i.e to be randomly distributed.

This ultimately implies that, entropy is a thermodynamic quantity that measures the degree of maximum disorder or randomness of a system.

The S.I unit used for the measurement of the degree of maximum order or randomness of a system is Joules per Kelvin (JK¯¹). An example of entropy is the mixing of ideal gases.

Generally, the entropy in an irreversible process always increases and as such the change in entropy has a positive value.

Hence, the entropy of the universe is always increasing because its energy flow is considered to be in a downward direction rather than upward i.e from a hot region to a cold region; making the energy to be evenly distributed.

Answer:

r =[tex]\frac{ 1 \pm \sqrt{ \frac{m}{M} } }{1 - \frac{m}{M} }[/tex]

Explanation:

Let's apply the universal gravitation law to the body (c), we use the indications 1 for the planet and 2 for the moon

          ∑ F = 0

           -F_{1c} + F_{2c} = 0

             F_{1c} = F_{2c}

let's write the force equations

             [tex]G \frac{m_c M}{r^2} = G \frac{m_c m}{(d-r)^2}[/tex]

where d is the distance between the planet and the moon.

              [tex]\frac{M}{r^2} = \frac{m}{(d-r)^2}[/tex]

             (d-r)² = [tex]\frac{m}{M} \ \ r^2[/tex]  

              d² - 2rd + r² = \frac{m}{M} \ \ r^2

              d² - 2rd + r² (1 - [tex]\frac{m}{M}[/tex]) = 0

              (1 - [tex]\frac{m}{M}[/tex])  r² - 2d r + d² = 0

we solve the second degree equation

              r = [2d ± [tex]\sqrt{ 4d^2 - 4 ( 1 - \frac{m}{M} ) }[/tex] ] / 2 (1- [tex]\frac{m}{M}[/tex])

              r = [2d ±  2d [tex]\sqrt{ \frac{m}{M} }[/tex]] / 2d (1- [tex]\frac{m}{M}[/tex])

              r =[tex]\frac{ 1 \pm \sqrt{ \frac{m}{M} } }{1 - \frac{m}{M} }[/tex]

there are two points for which the gravitational force is zero

The distance between object from planet will be "[tex]\frac{R}{[1+\sqrt{\frac{m}{M} } ]}[/tex]".

According to the question,

Let,

As we know,

→ [tex]Prerequisites-Gravitational \ force = \frac{GMm}{r^2}[/tex]

Now,

→ [tex]\frac{GMm_1}{x^2} = \frac{Gmm_1}{(R-x)^2}[/tex]

→ [tex]\frac{(R-x)^2}{x^2} = \frac{m}{M}[/tex]

→     [tex]\frac{R-x}{x} =\sqrt{\frac{m}{M} }[/tex]

→          [tex]x = \frac{R}{[1+ \sqrt{\frac{m}{M} } ]}[/tex]

Thus the answer above is appropriate.          

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Answer:

J = -0.522 m/s

Explanation:

Given that,

The mass of the baseball, m = 0.145 kg

Initial velocity, u = 14.5 m/s

Final velocity, v = 10.9 m/s

aWe need to find the direction of the impulse that the glass imparts to the baseball. Impulse is equal to the change in momentum such that,

[tex]J=m(v-u)[/tex]

Substitute all the values,

[tex]J=0.145\times (10.9-14.5)\\\\=-0.522\ kg-m/s[/tex]

The direction of impulse is opposite to the direction of velocity.

Answer:

534.8 meters

Explanation:

Use T=(2*pi*r)/v

560=(2*pi*r)/6

3360=2*pi*r

1680=pi*r

534.8 meters=radius

It takes 560s for a runner to complete one circular lap, moving at a speed of 6.00 m/s. The radius of a track is 534.7 m.

Mathematically, it can be calculated as follows :

distance = speed × time

The formula relating distance (d), speed (s), and time (t) is

d = st

First, Calculating the distance,

d = 560 s × 6 m·s⁻¹

  = 3360 m

When, Calculating the track radius,

The distance travelled is the circumference of a circle,

C = 2пr

r = 3360/2п

 = 534.7 m

The radius of the track is 534.7 m.

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Answer:

75.4

Explanation:

r= 2

h= 6

v= 22/7 *r*r*h

v= 75.42

Answer:

A Inertia

Explanation:

Answer:

v = 2.57 m / s

Explanation:

For this exercise let's use conservation of energy

starting point. When it is at an angle of 30º

          Em₀ = K + U = ½ m v₁² + m g y₁

final point. Lowest position

          Em_f = K = ½ m v²

as there is no friction, the energy is conserved

          Em₀ = Em_f

          ½ m v₁² + m g y₁ = ½ m v²

Let's find the height(y₁), which is the length of the thread minus the projection (L ') of the 30º angle

         cos 30 = L ’/ L

         L ’= L cos 30

         y₁ = L -L '

          y₁ = L- L cos 30

we substitute

          ½ m v₁² + m g L (1- cos 30) = ½ m v²

           v = [tex]\sqrt{ v_1^2 +2gL(1-cos30 )}[/tex]

let's calculate

           v = [tex]\sqrt{ 2^2 + 2 \ 9.8 \ 1.0 (1- cos 30)}[/tex]

           v = 2.57 m / s

Answer:

It went about 2 meters away

Explanation:

Answer:

v = 12.79 m/s

Explanation:

Given that,

The mass of a car, m = 1200 kg

Momentum of the car, p = 15 350 kg-m/s

We need to find the velocity of the car. We know that, the formula for the momentum of an object is given by :

p = mv

Where

v is the velocity of the bject

So,

[tex]v=\dfrac{p}{m}\\\\v=\dfrac{15350}{1200}\\\\v=12.79\ m/s[/tex]

So, the velocity of the car is 12.79 m/s.

Answer:

Δθ = 15747.37 rad.

Explanation:

       [tex]\omega_{f1} = \alpha * \Delta t = 2.38*e-3rad/s2*2.04e3s = 4.9 rad/sec (1)[/tex]

       [tex]\omega_{f1}^{2} = 2* \alpha *\Delta\theta (2)[/tex]

       [tex]\theta_{1} = \frac{\omega_{f1}^{2}}{2*\alpha } = \frac{(4.9rad/sec)^{2}}{2*2.38*e-3rad/sec2} = 5044.12 rad (3)[/tex]

       [tex]\theta_{2} =\omega_{f1} * \Delta_{t2} = 4.9 rad/s * 1.48*e3 s = 7252 rad (4)[/tex]

       [tex]\omega_{f2}^{2} - \omega_{o2}^{2} = 2* \alpha *\Delta\theta (5)[/tex]

      [tex]\theta_{3} = \frac{(\omega_{f2})^{2}- (\omega_{f1}) ^{2} }{2*\alpha } = \frac{(2.42rad/s^{2}) -(4.9rad/sec)^{2}}{2*(-2.63*e-3rad/sec2)} = 3451.25 rad (6)[/tex]

Answer:

C. Sigmund Freud . . . Carl Jung

Explanation:

edge 2021

C.Sigmund Freud and Carl Jung both heavily relied on dream analysis in their treatment of patients.

Freud is well-known for inventing and developing the approach of psychoanalysis; for articulating the psychoanalytic idea of motivation, intellectual infection, and the structure of the unconscious; and for influencing medical and popular conceptions of human nature by using positing both everyday and strange thought.

Sigmund Freud was an Austrian neurologist who's perhaps maximum known as the founding father of psychoanalysis. Freud advanced a fixed of therapeutic strategies centered on communication therapy that worried the use of techniques that include transference, loose affiliation, and dream interpretation.

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