A block of mass m = 8.40 kg, moving on a horizontal frictionless surface with a speed 4.20 m/s makes a perfectly elastic collision with a block of mass M at rest. After the collision, the 8.40 block recoils with a speed of 0.400 m/s. In the figure; the blocks are in contact for 0.200 s.

Answer:

1716.75 J

Explanation:

Step 1: First check what we are provided with. As per given question we have:

mass (m) = 70 kg, height (h) = 2.5 m and acceleration due to gravity (g) = 9.81 m/s².

Step 2: Check what we are asked to find out.

Work done = Change in Potential energy

The stuff required to solve this question is potential energy. Using the formula: P = mgh. Where P is Potential energy, m is mass, g is acceleration due to gravity and h is height.

Step 3: Substitute the known values in the above formula.

→ P = 70 × 2.5 × 9.81

→ P = 1716.75 J

Hence, the work done against the force of gravity is 1716.75 J.

We need evidence to be able to continually test theories about the composition and origin of our solar system and galaxies in order to verify the theories and to make laws.

A theory is an explanation for observations of the natural world that has been carried by using the scientific method, and which brings together many facts.

So we can conclude that we need evidence to test theories about the composition and origin of our solar system in order to verify the theories and to make laws.

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Increasing the temperature causes an increase in the average kinetic energy of the particles of a material.

The average kinetic energy of particles is the energy possessed by particles due to their constant motion.

The constant motion of particles occurs due to the energy acquired by the particles, when the temperature of the particles increases, the average kinetic energy increases which in turn increases the speed of the particles.

Thus, we can conclude that, increasing the temperature causes an increase in the average kinetic energy of the particles of a material.

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Answer:

Gravitational force is the force that attracts objects towards each other. Two factors that affect the gravitational force between objects are the mass of the two objects and the distance between

Explanation:

Gravity is what pulls us towards the Earth if we were to jump into the air, so it is the force that pulls things towards other things. The bigger the objects are the more gravity they have, so a planet has more gravity than say, an apple. Distance between objects also makes their gravity change, so the Earth's pull on the moon is different than the Earth's pull on the sun.

Hopefully this helps- let me know if you have any questions!

Describe gravitational force in your own words. Which two factors affect the gravitational force between two objects?

[tex]F = \frac{G m{ \tiny1}m{ \tiny2} }{ {r}^{2} } [/tex]

where

Answer:

wait in comments.................

Give the mathematical expression for coulomb's force if q1, q2 are the magnitude of charges and r is the distance between them.

F=K q1q2/r2

Answer:

p = rho× g × h

Explanation:

p: pressure

rho : density

g : gravity acceleration

h : height

Answer:

[tex]\displaystyle \sqrt{\frac{(5/2)\, (g - a)\, m\, R^{2}}{M^{2}\, a}}[/tex], assuming that the tension in the rope is the only tangential force on the sphere ([tex]g[/tex] denote the gravitational acceleration.)

Explanation:

The forces on the bucket are:

Since the weight of the bucket and the tension from the rope are in opposite directions, the magnitude of the net force would be:

[tex]\begin{aligned} \|\text{Net Force}\| =\; & \|\text{Weight}\| - \|\text{Tension}\| \end{aligned}[/tex].

The upward tension in the rope prevents the bucket from accelerating at [tex]g[/tex] (free fall.) Rather, the bucket is accelerating at an acceleration of only [tex]a[/tex]. The net force on the bucket would be thus [tex]m\, a[/tex].

Rearrange the equation for the net force on the bucket to find the magnitude of the tension in the rope would be:

[tex]\begin{aligned} & \|\text{Tension}\| \\ =\; & \|\text{Weight}\| - \|\text{Net Force}\| \\ =\; & m\, g - m\, a \\ =\; & (g - a)\, m\end{aligned}[/tex].

At a distance of [tex]R[/tex] from the center of the sphere, the tension in the rope [tex](g - a)\, m[/tex] would exert a torque of [tex](g - a)\, m\, R[/tex] on the sphere. If this tension is the only tangential force on this sphere, the net torque on the sphere would be [tex](g - a)\, m\, R\![/tex].

Let [tex]M[/tex] denote the mass of this sphere. The moment of inertia of this filled sphere would be [tex]I = (2/5)\, M^{2}[/tex].

Therefore, the magnitude of the angular acceleration of this sphere would be:

[tex]\begin{aligned}& \|\text{Angular Acceleration}\| \\ =\; & \frac{\|\text{Net Torque}\|}{(\text{Moment of Inertia})} \\ =\; & \frac{(g - a)\, m\, R}{(2/5)\, M^{2}} \end{aligned}[/tex].

The bucket is accelerating at a magnutide of [tex]a[/tex] downwards. The rope around the sphere need to unroll at an acceleration of the same magnitude, [tex]a\![/tex]. The tangential acceleration of the sphere at the surface would also need to be [tex]\! a[/tex].

Since the surface of the sphere is at a distance of [tex]R[/tex] from the center, the angular acceleration of this sphere would be [tex](a / R)[/tex].

Hence the equation:

[tex]\begin{aligned}& \frac{(g - a)\, m\, R^{2}}{(2/5)\, M^{2}} = \|\text{Angular Acceleration}\| = \frac{a}{R} \end{aligned}[/tex].

Solve this equation for [tex]M[/tex], the mass of this sphere:

[tex]\begin{aligned}& \frac{(g - a)\, m\, R^{2}}{(2/5)\, M^{2}} = \frac{a}{R} \end{aligned}[/tex].

[tex]\begin{aligned}M^{2} &= \frac{(g - a)\, m\, R^{2}}{(2/5)\, a} \\ &= \frac{(5/2)\, (g - a)\, m\, R^{2}}{a}\end{aligned}[/tex].

[tex]\begin{aligned}M&= \sqrt{\frac{(5/2)\, (g - a)\, m\, R^{2}}{a}}\end{aligned}[/tex].

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

[tex]X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}[/tex]

[tex]18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg[/tex]

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

[tex]W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN[/tex]

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

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(a) The time of flight be for a tennis ball launched by the ball machine is 0.19 s.

(b) The range of the tennis ball be is 1.01 m.

(c)  The final velocity of the ball with which it reaches the ground is 7.16 m/s.

The time of flight of the tennis ball is calculated as follows;

h = vt + ¹/₂gt²

1.2 = 5.3t + 0.5(9.8)t²

1.2 = 5.3t + 4.9t²

4.9t² + 5.3t - 1.2 = 0

a = 4.9, b = 5.3, c = 1.2

solve using quadratic formula

t = 0.19 s

Thus, the time of flight be for a tennis ball launched by the ball machine is 0.19 s.

The range of the tennis ball is calculated as follows;

R = vt

R = 5.3 x 0.19

R = 1.01 m

The final velocity of the ball with which it reaches the ground is calculated as follows;

vf = vo + gt

vf = 5.3 + 9.8(0.19)

vf = 7.16 m/s

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Now

[tex]\\ \sf\rightarrowtail Power=\dfrac{Work\:done}{Time\:taken}[/tex]

[tex]\\ \sf\rightarrowtail Power=\dfrac{250}{1}[/tex]

[tex]\\ \sf\rightarrowtail Power=250W[/tex]

Answer:

Newton

Explanation:

The SI unit for weight is newton.

Answer: I am so sorry i hadn't learned this Yet~!

Explanation:

The athlete can improve performance by building strength, coordination and balance.

This is an individual who is proficient in sports and other forms of physical exercise.

Improvement of performance based on the kinematic variable of speed and acceleration can be achieved by building strength, coordination and balance by performing plyometric exercises etc.

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[tex]\text{Given that, mass m = 70 kg and velocity v = 3 m/s}\\\\\text{Momentum,}~ p = mv = 70 \times 3 = 210~ kg ~ms^{-1}[/tex]

Hi there!

Begin by using Gauss' Law to find the electric field.

[tex]\oint {E \cdot} \, dA = \frac{Q_{encl}}{\epsilon_0}[/tex]

E = Electric field (N/C)
dA = differential area element

Q = enclosed charge (C)

ε₀ = Permittivity of free space (8.85 * 10⁻¹² C²/Nm²)

We can construct a large cylinder around the wire in order to determine the electric flux. The electric field lines will pass through the LATERAL surface area of the cylinder, so:
[tex]A = 2\pi rL[/tex]

Where 'L' is the length of the cylinder and 'r' is the distance from the capacitor.

The enclosed charge is equivalent to the charge per meter length (λ) multiplied by the length, so:
[tex]\oint {E \cdot} \, dA = \frac{\lambda L}{\epsilon_0}[/tex]

We can rewrite the dot product as EA (where cosθ = 1 since the normal vector points in the direction of the field).

A = the lateral surface area of a cylinder, so:

[tex]E * 2\pi rL = \frac{\lambda L}{\epsilon_0}[/tex]

Rearrange to solve for 'E'.

[tex]E = \frac{\lambda L }{2\pi r L \epsilon _0}\\\\E = \frac{\lambda }{2\pi r \epsilon_0}[/tex]

a)
Plug in the distance into 'r'.

[tex]E = \frac{\lambda }{2\pi r \epsilon_0} \\\\E = \frac{0.0000072}{2\pi * 5.5 * (8.85 * 10^{-12})} = \boxed{23542.18 \frac{N}{C}}[/tex]

b)
Repeat:
[tex]E = \frac{\lambda }{2\pi r \epsilon_0}\\\\E = \frac{0.0000072}{2\pi * 2.5 * (8.85 * 10^{-12})} = \boxed{51792.8 \frac{N}{C}}[/tex]

We can see that the distance from the wire is INVERSELY related to the electric field strength by a power of r⁻¹. The field strength DECREASES as the distance INCREASES.

Answer:

R1 + R2 = R = 12 for resistors in series - so R1 = R2 if they are identical

2 R1 = 12         and R1 = R2 = 6 ohms

1 / R = 1 / R1 + 1 / R2     for resistors in parallel

R = R1 * R2 / (R1 + R2) = 6 * 6 / (6 + 6) = 3

The equivalent resistance would be 3 ohms if connected in parallel

Answer:

16 reflection,incidence

Answer:

what is heat and transfer

If a person generates about 412005 joules of heat from the body,  the water temperature is mathematically given as

t=21.6296C

Question Parameter(s):

The person is submerged neck-deep into a tub containing 2124 kg of water at 20.9 °C

Generally, the equation for the Heat   is mathematically given as

Heat gained =Heat loess

Thereofore

mw*cw*(t-2160)=1.5*10^5

[tex]t=21.60+\frac{1.5*10^5}{mw*Cw}\\\\t=21.60+\frac{1.5*10^5}{1.2*10^3*4186}[/tex]

t=21.6296C

In conclusion, the tempreature

t=21.6296C

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The electric potential V(z) on the z-axis is :  V = [tex](\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z[/tex]

The magnitude of the electric field on the z axis is : E = kб 2[tex]\pi[/tex]( 1 - [z / √(z² + a² ) ] )

Given data :

V(z) =2kQ / a²(v(a² + z²) ) -z  

Considering a disk with radius R

Charge = dq

Also the distance from the edge to the point on the z-axis = √ [R² + z²].

The surface charge density of the disk ( б ) = dq / dA

Small element charge dq =  б( 2πR ) dr

dV  [tex]\frac{k.dq}{\sqrt{R^2+z^2} } \\\\= \frac{k(\alpha (2\pi R)dR}{\sqrt{R^2+z^2} }[/tex]  ----- ( 1 )

Integrating equation ( 1 ) over for full radius of a

∫dv = [tex]\int\limits^a_o {\frac{k(\alpha (2\pi R)dR)}{\sqrt{R^2+z^2} } } \,[/tex]

 V = [tex]\pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ][/tex]

     = [tex]\pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} } -z ][/tex]

Therefore the electric potential V(z) = [tex](\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z[/tex]

Also

The magnitude of the electric field on the z axis is : E = kб 2[tex]\pi[/tex]( 1 - [z / √(z² + a² ) ] )

Hence we can conclude that the answers to your question are as listed above.

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(a) The initial energy of the mass is 76.485 J.

(b) The compression of the spring in the absence of friction is 21.2 cm.

(c)  The compression of the spring in the presence of friction is 19.4 cm.

The initial energy of the mass is determined as follows;

E = K.E + P.E

E = ¹/₂mv² + mgh

where;

E = ¹/₂mv²  + mg x Lsinθ

P.E = ¹/₂(9.4)(2.5)² +  (9.4)(9.8)(0.87)(sin36)

P.E = 76.485 J

The compression of the spring in the absence of friction is calculated as follows;

Ux = E

¹/₂kx² = 76.485

kx² = 2(76.485)

x² = (2 x 76.485)/k

x = √(2 x 76.485)/k

x = √(2 x 76.485 / 3413.7)

x = 0.212 m

x = 21.2 cm

The compression of the spring in the presence of friction is determined by applying the principle of conservation of energy.

E - Fd = Ux

E - μmgd = ¹/₂kx²

76.11 - (0.27 x 9.4 x 9.8 x 0.48) = ¹/₂(3413)x²

64.17 = 1706.5x²

x² = 0.0376

x = √0.0376

x = 0.194

x = 19.4 cm

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The magnitude of magnitude force on the proton is 2.4 x 10⁻¹² N.

The magnitude of magnitude force on the proton is calculated as follows;

F = qvB sinθ

where;

Substitute the given parameters and solve for the magnetic force.

F = (1.6 x 10⁻¹⁹) x (3 x 10⁶) x (5) X(sin90)

F = 2.4 x 10⁻¹² N

Thus, the magnitude of magnitude force on the proton is 2.4 x 10⁻¹² N.

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Answer:

300 m/s

Explanation:

2d = vt

v = 2d/t

v = 2×90/.3

v=300 m/s

d = distance

t = time

v = velocity/speed of sound

Answer:

0

Explanation:

because no distance covered by body

In my own words, I would say that physics is an area of science that seeks to explain and understand the fundamental nature of the dynamics of objects, essentially defining how objects can interact, in space and in time.

When you rub a dry cloth across glass, it creates charged static electricity, which attracts little non-charged dust particles.