NEED HELP ASP
WILL GIVE POINTS OR WHATEVER

Answer:

C

Explanation:

The most commonly accepted theory of muscle growth states that microscopic tears within a muscle activate an inflammatory response process that causes dormant cells to build and regenerate muscle tissue; this is option c.

It has negative consequences on the muscle, and it is also known as "microtrauma," which occurs as a result of muscle contraction and is a normal part of the muscle repair and growth process. Here, the fibers in the muscle are damaged, and in response, the body initiates a repair process. This leads to muscle growth and increased strength and is helpful for muscle repair.

Hence, the most commonly accepted theory of muscle growth states that microscopic tears within a muscle activate an inflammatory response process that causes dormant cells to build and regenerate muscle tissue; this is option c.

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Answer:

True.

Explanation:

Exercises and yoga are important for the individual's health. The exercise makes the individual, fit, healthy and also reduces the chances of the deadly disease in human beings.

Different types of exercise can be performed by the individual. The aerobic exercises increase the amount of oxygen in the body. With the aerobic exercise, the exercise that can increase the strength, flexibility and total fitness must also be included.

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

  [tex]p = 14033.7 \ kg\cdot m/s[/tex]

b

  [tex]v = 1.22 \ m/s[/tex]

c

 [tex]m_2 = 9506.7 \ kg[/tex]  

Explanation:

From the question we are told that

     The mass of the coal car is  [tex]m_1 = 4527 \ kg[/tex]

     The velocity is  [tex]v_i = 3.1 \ m/s[/tex]

     The load of coal dropped is [tex]m_2 = 6952 \ kg[/tex]

Generally the momentum of the system before the coal is added is mathematically represented as

            [tex]p = m_j * v_i[/tex]

=>         [tex]p = 4527 * 3.1[/tex]

=>         [tex]p = 14033.7 \ kg\cdot m/s[/tex]

Generally from the law of momentum conservation is mathematically represented as

          [tex]m_1 * v_i + m_2 * v_2 = (m_1 + m_2 )v[/tex]

Here [tex]v_2[/tex] is the velocity of the load before it is dropped and the value is 0 m/s

So

 =>    [tex]v =\frac{m_1 * v_i }{m_1 + m_2}[/tex]

      [tex]v = \frac{4527 * 3.1 }{( 4527 + 6952 )}[/tex]

=>    [tex]v = 1.22 \ m/s[/tex]

When v =  1m/s  and  [tex]m_1[/tex] remain the same

Then

=>    [tex]1 =\frac{4527 * 3.1 }{4527 + m_2}[/tex]

=>    [tex]4527 + m_2 = 14033.7[/tex]

=>    [tex]m_2 = 9506.7 \ kg[/tex]  

Answer:

The average force ≅ 519.44 N.

Explanation:

Impulse = change in momentum of a body

i.e Ft = m(v - u)

where F is the force, t is the time, m is the mass of the body, v is the final velocity and u is the initial velocity.

m = 55.0 g (0.055 Kg), t = 0.00360 s, v = 34.0 m/s, since the ball was initially at rest; u = 0 m/s

So that,

F x 0.00360 = 0.055(34 - 0)

F x 0.00360 = 0.055 x 34

                    = 1.87

F = [tex]\frac{1.87}{0.0036}[/tex]

 = 519.4444

The average force exerted on the ball by the club is approximately 519.44 N.

Answer:

Explanation:

Frequency is oscillations per second.

So we have to find the Number of seconds she paced.

2 minutes = 2 X 60

               = 120 seconds

Therefore,

Her frequency = 10 / 120

                       = 1/12 Hertz

Answer:

Weather is the conditions (temperature, wind etc.) at a given time, like on that day. Climate, which is what his data would show, is the conditions over an extended period of time like the 3 months he collected data

Answer:

It allows electrons to flow from one part of the circuit to another

when it is closed.

Explanation:

hope this helps <3 !!

Answer: D

Explanation:

Answer:

20 K

Explanation:

It is given that,

The change in temperature is 20 C.

We need to find the change in thermodynamic temperature.

If teperauture T₁ = 0° C = 0+273 = 273 K

T₂ = 20° C = 20 + 273 = 293 K

The change in temperature,

[tex]\Delta T=T_2-T_1\\\\=293-273\\\\=20\ K[/tex]

So, the change in temperature of 20°C is equivalent to 20 K.

The change in thermodynamic temperature is equal to 20 Kelvin.

Given the following data:

To determine the change in thermodynamic temperature:

A thermodynamic temperature can be defined as an absolute measure of the average total internal energy possessed by a body or an object. Thus, thermodynamic temperature is typically measured in Kelvin (K).

Mathematically, the change in temperature of an object is given by the formula:

[tex]\theta = T_f - T_i[/tex]

Where:

Next, we would convert the values in degree Celsius to Kelvin:

Conversion:

0°C = 273 K

20°C = 273 + 20 = 293 K

For the change in thermodynamic temperature:

[tex]\theta = 293 - 273\\\\\theta =20\;K[/tex]

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Answer:

he is right he Is

Explanation:

step by step explanation

Answer:

0.5

Explanation:

Answer:

y becomes doubled.

Explanation:

If;

       y  = [tex]\frac{k}{x}[/tex]  

what is the state of y when x is halved;

  the given expression is an inverse relationship. When y increases, x is supposed to decrease and vice versa.

 if x is halved; x  = [tex]\frac{x}{2}[/tex]

           [tex]\frac{k}{\frac{x}{2} }[/tex]   = [tex]\frac{2k}{x}[/tex]  

Now compare :

      [tex]\frac{k}{x}[/tex]  :  [tex]\frac{2k}{x}[/tex]

we see that y becomes doubled

Answer:

0.130

Explanation:

From the given data, the coefficient of static friction for each trial are:

1. 0.053

2. 0.081

3. 0.118

4. 0.149

5. 0.180

6. 0.198

The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198

                                              = 0.779

So that;

the average coefficient of static friction = [tex]\frac{sum of coefficient of static friction}{number of trials}[/tex]

                                              = [tex]\frac{0.779}{6}[/tex]

                                              = 0.12983

The average coefficient of static friction is 0.130

The average coefficient of static friction is 0.13.

The coefficient of static friction is obtained using the formula; μ = F/R

Where;

F = force acting on the body

R = reaction

μ = coefficient of static friction

The average of measurements is given as; ∑summation of measurements/number of measurements

We can see from the question that there were 6 measurements of the coefficient of static friction. Hence, the average coefficient of static friction is obtained from;

0.053 + 0.081 + 0.118 + 0.149 + 0.180 +  0.198/6

= 0.13

The average  coefficient of static friction is 0.13

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The unit that should be used to measure energy when calculating specific heat capacity is Energy is in Joules ( J ).

The specific heat capacity should be determined in joules per kilogram degree-celsius ( J k g − 1 ∘ C − 1 ).

It is to be supplied for the substance with respect to mass and it increased the temperature.

Hence, The unit that should be used to measure energy when calculating specific heat capacity is Energy is in Joules ( J ).

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The unit used to measure energy when calculating specific heat capacity is the joule (J).

The joule (J), a unit used to quantify energy, is used to calculate specific heat capacity. In the International System of Units (SI), the joule serves as the default unit of energy.

It is described as the quantity of energy that is delivered when one newton of force is exerted across a one-meter distance.

Thus, the quantity of heat energy needed to increase the temperature of a particular substance by a specific amount is measured as specific heat capacity. J/kg°C, or joules per kilogramme per degree Celsius, is the unit of measurement.

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Answer:

Work done, W = 24.9001 J

Explanation:

Given that,

Force, F = (2xî + 4yĵ) Where x and y are in meters

This force acts on an object as the object moves in the x-direction from the origin to x = 4.99 m.

We need to find the work done by the force on the object. It is given by :

[tex]W=\int\limits^a_b {F dr} \\\\W=\int\limits^{4.99}_0 {(2xi+4yj)dx} \\\\W=\int\limits^{4.99}_0 {2xi\ dx} \\\\W=x^2|_0^{4.99}\\\\\text{Applying limits}\\\\W=(4.99)^2-0^2\\\\W=24.9001\ J[/tex]

So, the work done is 24.9001 J.

Answer:

Star Y is showing radial motion

Star Y is moving toward the Earth

Explanation:

Just answered this question on a quiz and got it right.

Star Y is showing radial motion

Star Y is moving towards earth.

Spectrum is defined as the arrangement of radiations emitted by atoms or molecules based on their characteristic wavelength and frequency.

Here,

The spectrum of the star Y is given and is compared to a reference hydrogen spectrum. The range of wavelengths of the star is given. The spectrum of the star shows that the wavelengths are shifted such that from longer wavelengths to shorter wavelength. This phenomenon of shift in wavelengths from higher to lower is known as blue shift. The star shows blue shift that means shifted to shorter wavelength or it can be said as shifted from lower frequency to higher frequency. As a result, it can be concluded that the star Y is moving towards the earth which implies Star Y is showing radial motion.

Hence,

The star Y is showing radial motion.

Star Y is moving towards the earth.

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Answer:

(a) The time taken for the plane to reach 50 km/hr is 811.97 s

(b) (i) the acceleration of the plane is 0.012 m/s²

    (ii) The time taken for the plane to reach 50 km/hr is 1172.85 s

Explanation:

Given;

mass of the freight train, m = 3.8 x 10⁷ kg

force applied to pull the train, F = 6.5 x 10⁵ N

a. How long does it take the train to go from rest to 50 km/hr?

Given;

initial velocity, u = 0

final velocity, v = 50 km/hr = 13.889 m/s

Apply Newton's second law of motion to determine the time traveled by the train at the given speed;

[tex]F = ma = \frac{m(v-u)}{t}\\\\t = \frac{m(v-u)}{F}\\\\t = \frac{3.8*10^7(13.889-0)}{6.5*10^5}\\\\t = 811.97 \ s[/tex]

(b)

(i) the acceleration of the plane

Given;

air resistance, = 2.0 x 10⁵ N

Apply Newton's second law of motion;

∑F = ma

6.5 x 10⁵ N - 2.0 x 10⁵ N = ma

4.5 x 10⁵ N = ma

[tex]a = \frac{4.5 * 10^5 }{m}\\\\a = \frac{4.5 * 10^5 }{3.8*10^7}\\\\a = 0.012 \ m/s^2[/tex]

(ii) time taken for the plane to reach 50 km/hr is given by;

[tex]F = \frac{m(v-u)}{t}\\\\t = \frac{m(v-u)}{F}\\\\t = \frac{3.8*10^7(13.889-0)}{4.5*10^5}\\\\t = 1172.85 \ s[/tex]

Answer:

Explanation:

a) centripetal acceleration is the acceleration of a body in a circular path. It is expressed as;

a = mv²/r

m is the mass of proton = 1.67×10^−27kg

v is the velocity = 3.00×10^8m/s

r is the radius

Since C = 2πr

7000m = 2πr

r = 7000/2π

r = 1114.08m

Substitute

a = 1.67×10^−27 (3.00×10^8)²/1114.08

a =  1.67×10^−27 * 9×10^16/1114.08

a = 15.03*10^-11/1114.08

a = 0.001346*10^-11

a = 1.346*10^-14m/s²

b) Force on the proton = mass * acceleration

Force = 1.67×10^−27kg * 1.346*10^-14

Force = 2.246*10^-41N

hence the force on the proton is 2.246*10^-41N

Answer:

Explanation:

To get the focal length, we will use the lens formula;

1/f = 1/u + 1/v

f is the focal length

u is the object distance

v is the image distance

Given

since the physicist's left eye is myopic, it will be corrected using concave lens and the image distance is negative.

u = 35cm

v = -2.0

1/f = 1/35-1/2

1/f = 2-35/70

1/f = -33/70

f = -70/33

f = -2.12 cm

f = -0.0212m

Power of a lend is the reciprocal of its focal length

Power of the lens = 1/f

P = 1/-0.0212

P = -47.17dioptres

The power of the lens is -47.17D

Answer:

The average coefficient of static friction = 0.130

Explanation:

The average coefficient of static friction = [tex]\frac{sum of coefficient of static friction}{number of trials}[/tex]

From the given question, the coefficient of static friction given are:

0.053, 0.081, 0.118, 0.149, 0.180, 0.198

Thus,

the sum of the given coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198

                                             = 0.779

The average coefficient of static friction = [tex]\frac{0.779}{6}[/tex]

                                             = 0.129833

                                             = 0.130

Therefore, the average coefficient of static friction is 0.130.

Answer:

N = 470 [N]

Explanation:

The normal force is defined as the reaction exerted by the surface where the body is located in the opposite direction to the weight component.

It can be easily calculated by means of the product of mass by gravitational acceleration.

N = m*g

where:

N = normal force [N] (units of Joules)

m = mass = 47 [kg]

g = gravity acceleration = 10 [N]

N = 47*10

N = 470 [N]

First of all, this isn't physics. But, here's the answer:

1 gram is a 1000 milligram and 250 milligram is 250 * 0.001 g which is 0.25 gram.

Since 1 kg is 1000 g, we divide 1000 by 0.25 and we get 1000 * 4 which is 4000.

There are [tex]4000[/tex] tablets inside the bottle.

Answer:

A.) acceleration= 55.6m/s^2

B.) acceleration of table= 5.0m/s^2

C.) More acceleration

Explanation:

A.) 100N/1.8kg= -55.6

B.) 100N÷20kg= 5

C.) Because since the table would have less mass, it would have had to accelerate more

Answer:

hindi ko gets

Explanation:

ang gulo naman nyan

Answer:

Option (c) is correct.

Explanation:

Acceleration of an object is given by the formula as follows :

[tex]a=\dfrac{v-u}{t}[/tex]

Where

u and v are initial and final velocity

t is time

(v-u) is also called the change in velocity

So, the acceleration of an object is equal to the rate of change of velocity. Hence, the correct option is (c) " Change in its velocity divided by the change in time".

Answer:

Poetic

Explanation:

I took the test

Answer:

Poetic

Explanation:

I got it right on the test

Answer:

Actually I know but I can't cause it will take time

Explanation:

so sorry

don't know but please mark me as brainliest please

Answer:

Magnitude of gravitational force between this planet and its moon: approximately [tex]1.37 \times 10^{20}\; \rm N[/tex].

Acceleration of this moon towards the planet: approximately [tex]5.49 \times 10^{-3}\; \rm m \cdot s^{-2}[/tex].

Acceleration of this planet towards its moon: approximately [tex]2.29 \times 10^{-5}\; \rm m \cdot s^{-2}[/tex].

Explanation:

Look up the gravitational constant, [tex]G[/tex]:

[tex]G \approx 6.67 \times 10^{-11}\; \rm m^3\cdot kg^{-1} \cdot s^{-2}[/tex].

Assume that both this planet and its moon are spheres of uniform density. When studying the gravitational interaction between this planet and its moon, this assumption allows them to be considered as two point masses.

The formula for the size of gravitational force between two point masses [tex]m_1[/tex] and [tex]m_2[/tex] with a distance of [tex]r[/tex] in between is:

[tex]\displaystyle F = \frac{G \cdot m_1 \cdot m_2}{r^2}[/tex],

where [tex]G[/tex] is the gravitational constant.

Let [tex]m_1[/tex] and [tex]m_2[/tex] denote the mass of this planet and its moon, respectively.

Calculate the size of gravitational force between this planet and its moon:

[tex]\begin{aligned} F &= \frac{G \cdot m_1 \cdot m_2}{r^2} \\ &\approx \frac{6.67 \times 10^{-11}\; \rm m^3 \cdot kg^{-1} \cdot s^{-2} \times 6.00 \times 10^{24}\; \rm kg \times 2.50 \times 10^{22}\; \rm kg}{{\left(2.70 \times 10^{8}\; \rm m\right)}^2} \\ &\approx 1.37 \times 10^{20}\; \rm N\end{aligned}[/tex].

Assume that other than the gravitational force between this planet and its moon, all other forces (e.g., gravitational force between this planet and the star) are negligible. The magnitude of the net force on the planet and on the moon should both be approximately [tex]1.37 \times 10^{20}\; \rm N[/tex].

Apply Newton's Second Law of motion to find the acceleration of this planet and its moon:

[tex]\displaystyle \text{acceleration} = \frac{\text{net force}}{\text{mass}}[/tex].

For this moon:

[tex]\displaystyle \frac{1.37 \times 10^{20}\; \rm N}{2.50 \times 10^{22}\; \rm kg} \approx 5.49\times 10^{-3}\; \rm m \cdot s^{-2}[/tex].

For this planet:

[tex]\displaystyle \frac{1.37 \times 10^{20}\; \rm N}{6.00 \times 10^{24}\; \rm kg} \approx 2.29 \times 10^{-5}\; \rm m \cdot s^{-2}[/tex].