A sleigh is being pulled horizontally by a train of horses at a constant speed of 6.38 m/s. The magnitude of the normal force exerted by the snow-covered ground on the sleigh is 7.50 ✕ 103 N.
(a) If the coefficient of kinetic friction between the sleigh and the ground is 0.26, what is the magnitude of the kinetic friction force experienced by the sleigh?
N

(b) If the only other horizontal force exerted on the sleigh is due to the horses pulling the sleigh, what must be the magnitude of this force?
N

Answer:

2.096m/s

Explanation:

The speed of this soccer ball can be calculated using the formula;

Speed = distance/time

According to this question, the distance of the ball before it hits the tree is 6.5m, the time it takes is 3.1s, hence;

Speed = 6.5/3.1

Speed of the ball = 2.096m/s

Therefore, the speed of the ball before hitting the tree is 2.096m/s

Answer:

the intellectual and practical activity encompassing the systematic study of the structure and behaviour of the physical and natural world through observation and experiment.

Explanation:

Answer:

6.7 m/s

Explanation:

Given:

Δx = 5 m

v₀ = 5 m/s

a = 2 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (5 m/s)² + 2 (2 m/s²) (5 m)

v = 6.7 m/s

Answer:

they were in two places in flint and Birmingham and in Birmingham it is hot but flint of cold the Simi is they both have Sunday school for Joetta

Explanation:

use in your own words teachers know when your not trust me.

Answer:

The speed as it reaches the bottom is  24m/s

Explanation:

Given parameters:

   Height of hill  = (3 x 9.8)m = 29.4m

Unknown:

Final speed as it reaches the bottom = ?

Solution:

 To solve this problem, we apply;

    v² = u² + 2gh

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity = 9.8m/s²

h is the height

note, the initial speed of the body is 0;

Input the parameters and solve;

    v² = 0² + 2 x 9.8 x 29.4

   v² = 576.24

   v = √576.24  = 24m/s

The speed as it reaches the bottom is  24m/s

Answer: yes it can

Explanation:

Answer:

Yes you can. You need to eat healthy and have the proper diet and always exercise and build muscles, this can slow down aging, and most importantly, we must always be calm. However there is no way to stop aging for that’s part of life.

Explanation:

Answer:

KE= 1/2mv²

Explanation:

The kinetic energy of a body is the energy possessed by virtue of the body in motion

Given the parameters

m which is the mass of the body

v which is the velocity of the body too

K.E = kinetic energy

The expression for the kinetic energy of a body is given as

KE= 1/2mv²

Answer:

Acceleration = m/s²

Explanation:

T= Newtons compared to the weight W = Newtons for the hanging mass. If the weight of the hanging mass is less than the frictional resistance force acting on the mass on the table, then the acceleration will be zero.

Answer:

Thermal Energy

Explanation:

Answer:

(a). Mario's speed in m/s = 5.2 × 10^-3 m/s.

Bowser Jr.'s speeds in m/s = 3.92 × 10^-3 m/s.

(b). 27001.2 seconds(s)..

(c). 141 metre(m).

Explanation:

So, the following data or parameters or information was given in the question above. These informations are going to help us in solving this question or problem;

=>" Bowser Jr. then passes Mario at his top speed = 30 blocks/h.

=> " Mario quickly accelerates and reaches his top speed of 40 blocks/h in order to catch back up and pass Bowser Jr., but by this point he has opened up a 75 block head start down the track."

=> "Mario world measure distance using the units of blocks, with 1 block = 0.47 m"

Therefore, the solution is given below;

(1). For the first part, we are to determine or calculate Mario and Bowser Jr.'s speeds in m/s.

Therefore, Mario's speed in m/s = 40 × 0.47) ÷ 3600 = 5.2 × 10^-3 m/s.

Also, Bowser Jr.'s speeds in m/s = ( 30 × 0.47) ÷ 3600 = 3.92 × 10^3 m/s.

(2). So, the next thing to do now.is to determine or calculate how long it took for Mario to catch Bowser Jr.

Thus, the time it took for Mario to catch Bowser Jr. Can be related as below;

[ ( 5.2 × 10^-3 m/s) - (3.92 × 10^-3 m/s) × (time,t taken for Mario to catch Bowser Jr.) = 75 × 0.47.

Therefore, the time it took for Mario to catch Bowser Jr. = 27001.2 seconds.

(3). Now, we calculate How far down the track Mario from the point at which he reaches his top speed.

The distance = 5.2 × 10^-3 m/s × 27001.2m = 141m

Answer:

The period of that same pendulum on the moon is 12.0 seconds.

Explanation:

To determine the period of that same pendulum on the moon,

First, we will determine the value of g (which is a measure of the strength of Earth's gravity) on the Moon. Let the value of g on the Moon be [tex]g_{M}[/tex].

From the question, the strength of earth’s gravity is only 1/6th of the normal value. The normal value of g is 9.8 m/s²

∴ [tex]g_{M}[/tex] = [tex]\frac{1}{6} \times 9.8 m/s^{2}[/tex]

[tex]g_{M}[/tex] = 1.63 m/s²

From the question, T=2π√L/g

[tex]T = 2\pi \sqrt{\frac{L}{g} }[/tex]

We can write that,

[tex]T_{E} = 2\pi \sqrt{\frac{L}{g_{E} } }[/tex] .......... (1)

Where [tex]T_{E}[/tex] is the period of the pendulum on Earth and [tex]g_{E}[/tex] is the measure of the strength of Earth's gravity

and

[tex]T_{M} = 2\pi \sqrt{\frac{L}{g_{M} } }[/tex] .......... (2)

Where [tex]T_{M}[/tex] is the period of the pendulum on Moon and [tex]g_{M}[/tex] is the measure of the strength of Earth's gravity on the Moon.

Since we are to determine the period of the same pendulum on the moon, then, [tex]2\pi[/tex] and [tex]L[/tex] are constants.

Dividing equation (1) by (2), we get

[tex]\frac{T_{E} }{T_{M} } = \sqrt{\frac{g_{M} }{g_{E} } }[/tex]

From the question,

[tex]T_{E} = 4.9secs[/tex]

[tex]g_{E}[/tex] = 9.8 m/s²

[tex]g_{M}[/tex] = 1.63 m/s²

[tex]T_{M}[/tex] = ??

From,

[tex]\frac{T_{E} }{T_{M} } = \sqrt{\frac{g_{M} }{g_{E} } }[/tex]

[tex]\frac{4.9}{T_{M} } = \sqrt{\frac{1.63}{9.8} }[/tex]

[tex]\frac{4.9}{T_{M} } = 0.40783[/tex]

[tex]T_{M} =\frac{4.9}{0.40783 }[/tex]

[tex]T_{M} = 12.01 secs[/tex]

∴ [tex]T_{M} = 12.0secs[/tex]

Hence, the period of that same pendulum on the moon is 12.0 seconds.

Answer:

The period of that same pendulum on the moon is 12.0 s

Explanation:

Given;

period of a pendulum’s swinging, T=2π√L/g

the strength of earth’s gravity on moon, g₂ = ¹/₆(g₁)

period of pendulum on Earth, T₁ = 4.9 s

period of pendulum on moon, T₂ = ?

The length of the pendulum is constant, make it the subject of the formula;

[tex]T = 2\pi \sqrt{\frac{L}{g} }\\\\\frac{T}{2\pi} = \sqrt{\frac{L}{g}}\\\\(\frac{T}{2\pi} )^2 =\frac{L}{g}\\\\\frac{T^2}{4\pi^2} = \frac{L}{g}\\\\ L = \frac{gT^2}{4\pi^2}\\\\L_1 = L_2\\\\\frac{g_1T_1^2}{4\pi^2}= \frac{g_2T_2^2}{4\pi^2}\\\\g_1T_1^2 = g_2T_2^2\\\\T_2^2 = \frac{g_1T_1^2}{g_2} \\\\T_2 = \sqrt{\frac{g_1T_1^2}{g_2}}\\\\ T_2 = \sqrt{\frac{g_1T_1^2}{g_1/6}}\\\\ T_2 = \sqrt{\frac{6*g_1T_1^2}{g_1}}\\\\T_2 = \sqrt{6T_1^2}\\\\ T_2 = T_1\sqrt{6} \\\\T_2 = (4.9)\sqrt{6}\\\\ T_2 = 12.0 \ s[/tex]

Therefore, the period of that same pendulum on the moon is 12.0 s

Answer:

The current in the loop is 10.5 A.

Explanation:

Given that,

Radius = 9.4 cm

Magnetic field = 0.7 G

Angle = 70°

We know that,

The magnetic field due to the current in a loop is

[tex]B_{I}=\dfrac{\mu_{0}NI}{2r}[/tex]

The magnetic field due to the current is equal to the magnetic field of earth.

[tex]B_{E}=B_{I}=\dfrac{\mu_{0}NI}{2r}[/tex]

We need to calculate the current in the loop

Using formula of magnetic field

[tex]B=\dfrac{\mu_{0}NI}{2r}[/tex]

[tex]I=\dfrac{2rB}{\mu_{0}N}[/tex]

Put the value into the formula

[tex]I=\dfrac{2\times9.4\times10^{-2}\times0.7\times10^{-4}}{4\pi\times10^{-7}\times1}[/tex]

[tex]I=10.5\ A[/tex]

Hence, The current in the loop is 10.5 A.

Answer:

Its A

Explanation:

Just did the quiz

The pair of objects that should be strongly attracted is option A. A positively charged particle and a negatively charged particle.

When there is a positively charged particle  & the negatively charged particle so due to this it should be strongly attracted.  Also, when there are two positively charged particles, or a negative one or a include negative one or neutral one so it should not be strongly attracted. Therefore, the option A is correct.

learn more about particles here: https://brainly.com/question/12734041

This question is incomplete, the complete question is;

A gong makes a loud noise when struck. The noise gradually gets less and less loud until it fades below the sensitivity of the human ear. The simplest model of how the gong produces the sound we hear treats the gong as a damped harmonic oscillator. The tone we hear is related to the frequency f of the oscillation, and its loudness is proportional to the energy of the oscillation.

If the loudness drops to 90 % of its original value in 5.0 s , what is the time constant of the damped oscillation

Answer: the time constant of the damped oscillation is 47.44s

Explanation:

Given that;

t = 5.0s

Lets say Ao is the amplitude of initial loudness and later A(t) = 0.9 Ao

EXPRESSION for amplitude is  A(t) = Ao e^-t / T

t is time while T is time constant

so

0.9Ao = Ao e^-t / T  

0.9 = e^ -t/T

So we take the natural log of both the sides

ln (0.9) = -t/T

-0.1054 =  -t/T

0.1054 =  t/T

WE now substitute our value of t

0.1054 =  t/T

0.1054T =  5.0

T = 5 / 0.1054

T = 47.44s

therefore the time constant of the damped oscillation is 47.44s

Answer:

The maximum torque on the loop is 395.80 N.m.

Explanation:

Given;

number of turns of the wire, N = 150 turns

length of the square loop, L = 18.0 cm = 0.18 m

current in the wire, I = 50.9 A

Magnetic field, B = 1.6 T

Maximum torque on the loop is given by;

τ = NIAB

τ = (150)(50.9)(0.18²)(1.6)

τ = 395.80 N.m

Therefore, the maximum torque on the loop is 395.80 N.m.

Answer: Spring constant = 472N/m

Explanation:

The change in  gravitational potential energy by the spring is given as = mgh

where m= 7.0 g = 7 X 10 -3kg

g= 9.8m/s

h= 22m

Gravitational potential energy= mgh

= 7.0 x  10^-3 X 9.8  x 22 = 1.5092 J

Remember that  change in  gravitational potential energy by the spring =elastic potential energy

Therefore,  Potential energy P. E = 1/2 K x²

where K= CONSTANT

x= 8.0

2 X 1.5092 J / (8.0 X 10^-2)²= 471.625 ROUNDED TO 472 N/m

Answer: D 160kg object moving 126 m/s

Explanation:

An object having a mass of 162 kg and moving with a velocity of 124 m/sec will require the greatest amount of force to change its motion. The correct option is D.

Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

If the object has to stop, the final velocity must be zero. If the time is constant, the amount of force only depends on the mass and the velocity at which the body is moving.

The amount of force on the object depends on the momentum of the body.

The momentum of the body is;

P = mv

Object D will require the greatest amount of force to change its motion. Because the momentum of the body for option D is the greatest.

Hence, the correct option is D.

Learn more about the Force, here;

https://brainly.com/question/26115859

#SPJ2

Explanation:

When sound travels in a given "medium", it would touch the surface of another "medium" and will bounce back in some other direction, this occurrence is called the reflection of sound.

Answer:

5.0 m/s south

Hope this Helps!

Answer:

5.0 m/s south

Explanation:

Ohm's law shows the relationship between voltage, current, and the resistance of a energy bond. The formula for Ohm's law is:

voltage = current x resistance

This formula tells you that current and resistance is the voltage of an energy bond.

Hope this helps you!

Answer: What is Ohm’s Law?

Ohm's Law is a formula used to calculate the relationship between voltage, current and resistance in an electrical circuit.

Ohm's Law (E = IR) is as fundamentally important as Einstein's Relativity equation (E = mc²) is to physicists.

E = I × R

It means voltage = current × resistance, or volts = amps × ohms, or V = A × Ω.

Resistance cannot be measured in an operating circuit, so Ohm's Law is especially useful when it needs to be calculated. Rather than shutting off the circuit to measure resistance, a technician can determine R using the above variation of Ohm's Law.

Plz mark brainliest:)

Answer:

We are given:

V1 = 5.5L          T1 = -38 C   or   235 k

V2 = 1.3L           T2 = T

From the gas equation:

PV = nRT

Since the pressure (P) , number of moles (n) and the universal gas constant (R) are constants, we can write the same equation as:

V / T = k  (where k is a constant)

so a bit more insight, since the values noted above are constant, when multiplied by each other, they will provide us with a constant number irrespective of the value of the variables

Changing the variables for the first case:

V1 / T1 = k   (where k is the same constant) ----------------(1)

Similarly,

V2 / T2 = k  (again, k has the same value)----------------(2)

From (1) and (2):

k is the common value

V1 / T1 = V2 / T2

Replacing the variables

5.5 / 235 = 1.3 / T

T = 1.3 * 235 / 5.5

T = 55.54 k

Therefore, at 55.54 K the gas will have a volume of 1.3L

v = u + a t

where u = initial velocity (25 m/s), v = final velocity (0), a = acceleration, and t = time (45 s). So

0 = 25 m/s + a (45 s)

a = (-25 m/s) / (45 s)

a ≈ -0.56 m/s²

Answer:

The net force on an object is the total force applied on the object after adding  up all the forces

In the given diagram,

we can see that the 2 forces of 4N and 4N will cancel each other out since they are equal and in the opposite direction

Now, we are left with a force of 2N and 10N,

the net force will be the difference of these forces:

Net force = 10N - 2N

Net force = 8N downwards

Another way to do it:

The two 4N forces will be cancelled out,

and we are left with a 2N and a 10N force

(notice how we cancelled equal and opposite forces for the 4N)

We can divide the 10N force into (2N + 8N)

Since the 2N forces are equal and opposite, they will be cancelled out

and we will be left with a net force of 8N downwards

Answer: I'm pretty sure the answer is B

Explanation: I'll check after the test

:Okay yes the answer is 100% B

The direction of the magnetic field shown by the magnetic field lines around a magnet is from north to its south pole of magnet. Option B is correct.

Magnetic Field Lines:

These are the imaginary lines that are used to represent the strength of the magnetic field.

Therefore, the direction of the magnetic field shown by the magnetic field lines around a magnet is from north to its south pole of magnet.

To know more about Magnetic Field Lines,

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