३.रात में घूमने वाला write one word substitute ​

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:

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:

If you have two batteries and they have precisely the same voltage then placing one backwards will effectively cancel out the voltages and no current will flow. However, batteries aren't like that. The slightest difference in voltages mean that current will flow.

Explanation:

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

Answer:

work done= force × displacement

=800×5

=4000J

Explanation:

The amount of work done is the result of the magnitude of force applied and the displacement of the body due to the force applied. Therefore, work done is defined as the product of the applied force and the displacement of the body.

Answer:

they were fast ⛷⛷

Answer: if weight affects how fast they go?

Explanation:

Answer:

How can we change the speed of a toy car on a racetrack to describe the car’s motion?

Explanation:

thats the sample respond

Answer:

75.4

Explanation:

r= 2

h= 6

v= 22/7 *r*r*h

v= 75.42

I don't know the answer but I just want points sorry

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]

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

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: 2000 J

Explanation: work W = F s

Answer:

8.45 m

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 90 Kg

Initial velocity (u) = 13 m/s

Final velocity (v) = 0 m/s

Height (h) =?

NOTE: Acceleration due to gravity (g) = 10 m/s²

The height of the hill can be obtained as follow:

v² = u² – 2gh (since the cart is going against gravity)

0² = 13² – (2 × 10 × h)

0 = 169 – 20h

Rearrange

20h = 169

Divide both side by 20

h = 169/20

h = 8.45 m

Therefore, the height of the hill is 8.45 m

Answer:

Regular reflection

Explanation:

Regular reflection occurs when light reflects off a very smooth surface and forms a clear image.

i hope this helps a bit.

According to the context, a sharp image is formed when light reflects from a regular reflection.

It is reflection without diffusion that obeys the laws of geometrical optics, as in mirrors.

This reflection of light happens when the angles that the two rays determine with the surface are equal.

Therefore, we can conclude that according to the context, a sharp image is formed when light reflects from a regular reflection.

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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:

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:

1) 1000Nm

2)  95,625Nm

3) 1.05%

Explanation:

Mechanical Advantage is the ratio of the load  to the effort applied to an object.

MA = Load/Effort

1) Workdone on the load = Force(Load) * distance covered by the load

Workdone on the load = 500N * 2m

Workdone on the load = 1000Nm

2)  work done by the effort = Effort * distance moves d by effort

work done by the effort = 2125 * 45

work done by the effort = 95,625Nm

3) Efficiency = Workdone on the load/ work done by the effort * 100

Efficiency = 1000/95625 * 100

Efficiency = 1.05%

Hence the efficiency of the system is 1.05%

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:

(a) The volume of water is 100 cm³

(b) The volume of the rock is 20 cm³

(c) The density of the rock is 30 g/cm³

Explanation:

The given parameters of the perspex box are;

The area of the base of the box, A = 10 cm²

The initial level of water in the box, h₁ = 10 cm

The mass of the rock placed in the box, m = 600 g

The final level of water in the box, h₂ = 12 cm

(a) The volume of water in the box, 'V', is given as follows;

V = A × h₁

∴ The volume of water in the box, V = 10 cm² × 10 cm = 100 cm³

The volume of water in the box, V = 100 cm³

(b) When the rock is placed in the box the total volume, [tex]V_T[/tex], is given by the sum of the rock, [tex]V_r[/tex], and the  water, V, is given as follows;

[tex]V_T[/tex] = [tex]V_r[/tex] + V

[tex]V_T[/tex] = A × h₂

∴ [tex]V_T[/tex] = 10 cm² × 12 cm = 120 cm³

The total volume, [tex]V_T[/tex] = 120 cm³

The volume of the rock, [tex]V_r[/tex] = [tex]V_T[/tex] - V

∴ [tex]V_r[/tex] = 120 cm³ - 100 cm³ = 20 cm³

The volume of the rock, [tex]V_r[/tex] = 20 cm³

(c) The density of the rock, ρ = (Mass of the rock, m)/(The volume of the rock)

∴ The density of the rock, ρ = 600 g/(20 cm³) = 30 g/cm³

Answer:

athletic

Explanation:

because internet system has been down since we were in few days

Answer:

It's plastic.

trust me it's plastic, i've rad it somewhere.

All of them have something that's not like the others.

-- Rubber is the only one on the list that has two repeated letters.

-- Plastic is the only one on the list thagt has no repeated letters.

-- Plastic is the only one on the list that has no 'r' in its name.

-- Copper is the only one on the list that is an element, not a compound.

-- Copper is the only good electrical conductor on the list.

-- Plastic is the only one on the list with more than six letters in its name.

-- Rubber is the only one on the list with no 'p' in its name.

-- Plastic is the only one on the list that doesn't end in "-er".

Answer:

c.

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

Answer:

B. in the direction of the force

Explanation:

Sana nakatulong

Answer:

37.5 m/s

Explanation:

Using,

Formula

v = ωr....................... Equation 1

Where ω = instantaneous angular velocity, v = instantaneous linear velocity, r = radius or distance from the sweet spot of the bat to the axis of rotation.

From the question,

Given: ω = 30 rad/s, r = 1.25 m

Substitute these values into equation 1

v = 30(1.25)

v = 37.5 m/s.

Hence the instantaneous linear velocity of the sweet spot at the instant of ball contact is 37.5 m/s