Help me please I have other ones like this too on my page please help!

Answer:

B

Explanation:

Where it shows on the map is B

Answer:

a) 6636 km

b) 0.0154

Explanation:

The height above the earth at its furthest point is 368 km

The height above the earth at its closest point is 164 km

Radius of the Earth is 6370 km

The distance of the satellite from the center of the earth to the furthest point is 6370 + 368 km = 6738 km

The distance of the satellite from the center of the earth to the closest point is 6370 + 164 = 6534 km

If we add together the sum of the distance of the satellite from the furthest and its closest distance, it is equal to the 2 major semi axis.

Basically,

2a = R + r

a = (R + r) / 2

a = (6738 + 6534) / 2

a = 13272 / 2

a = 6636 km

Eccentricity, e = (a - r) / a

Eccentricity, e = (6636 - 6534) / 6636

Eccentricity, e = 102 / 6636

Eccentricity, e = 0.0154

Answer:

Centripetal acceleration is defined as the property of the motion of an object, traversing a circular path. Any object that is moving in a circle and has an acceleration vector pointed towards the center of that circle is known as Centripetal acceleration. ... Centripetal means towards the center.

Answer:

rotational kinetic energy is 1156.81 J

Explanation:

Given the data in the question;

first we find the moment of inertia of the rod as axis passes through the middle or center;

[tex]I = \frac{1}{12}ml^2[/tex]

where m is the mass of the road( 3.0 kg)

[tex]l[/tex] is the length of the rod ( 2.6 m )

so we substitute

[tex]I = \frac{1}{12}[/tex] × 3 × (2.6)²

[tex]I[/tex] =   1.69 kg.m²

now, to calculate the rotational kinetic energy,

kinetic energy due to rotation is;

KE[tex]_R[/tex] = [tex]\frac{1}{2}Iw^2[/tex]

where [tex]I[/tex] is moment of inertia( 1.69 kg.m² ),

w is angular frequency ( 37 radians/s )

we substitute

KE[tex]_R[/tex] = [tex]\frac{1}{2}[/tex] × 1.69 × ( 37 radians/s )²

KE[tex]_R[/tex] = 1156.81 J

Therefore, rotational kinetic energy is 1156.81 J

The rotational kinetic energy will be 1156.81 J. It is half of the product of the moment of inertia and angular speed.

The rotational kinetic energy is the kinetic energy generated by an object's rotation and It is a component of its overall kinetic energy.

The given data in the problem is;

m is the mass = 3.0 kg

l is the length = 2.6 m

[tex]\rm \omega[/tex] is the angular speed = 37 radians/s

v is the speed = 13 m/s

[tex]\rm E_k[/tex] is the  rotational kinetic energy=?

The momentum of inertia is found by;

[tex]\rm I = \frac{1}{12} ml^2 \\\\ \rm I = \frac{1}{12} \times 3 \times (2.6)^2 \\\\\ \rm I = 1.69 \kg.m^2[/tex]

The rotational kinetic energy is found by;

[tex]\rm E_k= \frac{1}{2}I (\omega)^2 \\\\ \rm E_k= \frac{1}{2}\times 1.69 (37)^2 \\\\ \rm E_k=1156.81 \ J[/tex]

Hence the rotational kinetic energy will be 1156.81 J.

To learn more about the rotational kinetic energy refer to the link;

https://brainly.com/question/19305456

Answer:

Δx = 2*d

Explanation:

        [tex]k*\Delta x = m*g (1)[/tex]

       [tex]F = k_{eff} * \Delta x_{eff} = m*g (2)[/tex]

       [tex]\Delta x_{1} = \frac{F}{\ k_{1} } (3)[/tex]

       [tex]\Delta x_{} = \frac{F}{\ k_{1} } + \frac{F}{\ k_{2} } (4)[/tex]

       [tex]\frac{F}{k_{eff} } = \frac{F}{\ k_{1} } + \frac{F}{\ k_{2} } (5)[/tex]

      Since k₁ = k₂ = k, we can find keff, as follows:

      [tex]k_{eff} = \frac{k^{2} }{2*k} = \frac{k}{2} (6)[/tex]

     [tex]F = \frac{k}{2} * \Delta x_{eff} = k*\Delta x = m*g (7)[/tex]

       [tex]\Delta x_{eff} = \frac{2*k*\Delta x}{k} = 2* \Delta x = 2*d (8)[/tex]

Answer:

τ = RC = 2.17 s

Explanation:

       [tex]V_{C} =V_{o} * e^{-t/RC} (1)[/tex]

       [tex]V_{C} = 1.00 V\\V_{o} = 100V\\t = 10.0 s[/tex]

      [tex]\frac{1.00V}{100 V} = e^{-10s/RC} (2)[/tex]

       [tex]R*C= \frac{-10s}{ln 0.01} = 2.17s (3)[/tex]

Answer: Because of the amount of energy

Explanation: Heat is an energy, and small amounts of it does not effect large areas, whereas a very large blast of heat energy can greatly effect the air

Answer:

1) Linear motion

2) Rotational motion

3) Rotational Motion

4) Random Motion ( The ball can be rolling in any direction)

5) Linear motion

The types of motion in each activity include the following:

This involves an object or a body changing position over time. There are

different types of motion with different examples as can be seen above in

this scenario.

Read more about Motion here https://brainly.com/question/453639

Answer:the answer is galaxy

Explanation:

Answer:

true

Explanation:

i think its true beacuse water can cause electricity

pls let me know if corrct

if yes pls mark brainlest

Answer:

Acceleration = Δv/Δt or change in velocity over change in time

Explanation:

Answer:

μ = mg/kx

Explanation:

Since the bock does not slip, the frictional force equals the weight of the block. So, F = mg. Now, the frictional force, F = μN where μ = coefficient of static friction and N = Normal force.

Now, the normal force equals the spring force F' = kx where k = spring constant and x = compression of spring.

N = F' = kx

So, F = μN = μkx

μkx = mg

So, μ = mg/kx

Answer:

while ice is made by water again it melts and becomes water. water is colourless and odourless and has no taste but ice is only cold and hard. water is used for drinking and other things. but is for freshness and it never flows

Explanation:

so ice reflect more energy compared to water

Answer:

v' = -18 m/s

Explanation:

       [tex]p_{o} = p_{f} (1)[/tex]

       [tex]m_{b} * v_{b} -M_{c}*V_{c} = m_{b} * 18 m/s + (-M_{c}* 20 m/s) (2)[/tex]

        [tex]-(m_{b} * v'_{b}) + M_{c}*V'_{c} (3)[/tex]

       [tex]-(m_{b} * v'_{b}) +(- M_{c}*V_{c}) = -(m_{b} * v'_{b}) + (-M_{c} * 20 m/s) (4)[/tex]

       [tex]m_{b} * 18 m/s + (-M_{c}* 20 m/s) = -(m_{b} * v'_{b}) + (-M_{c} * 20 m/s) (5)[/tex]

Answer:

a)   t = 19.6 s, b) fr = 1.274 10⁴ N

Explanation:

This is a Newton's second law problem

Y Axis

for the cabin

        N₁-W₁ = 0

        N₁ = W₁

for the trailer

        N₂- W₂ = 0

        N₂ = W₂

X axis

for the cabin plus trailer, where friction is only in the cabin

         fr = (m₁ + m₂) a

the friction force equation is

        fr = μ N

we substitute

       μ N₁ = (m₁ + m₂) a

        μ m₁ g = (m₁ + m₂) a

        a = μ g    [tex]\frac{m_1}{m_1 + m_2}[/tex]

let's calculate

         a = 0.65 9.8    [tex]\frac{2000}{2000+8000}[/tex]

         a = 1,274 m / s²

a) to find the stopping distance we can use kinematics

         Let's slow down the sI system

         v₀ = 90 km / h (1000 m / 1km) (1h / 3600s) = 25 m / s

         v = v₀ - a t

when it is stopped its speed is zero

           0 = v₀ - at

           t = v₀ / a

           t = 25 / 1.274

           t = 19.6 s

b) the friction force is

           fr = 0.65 2000 9.8

          fr = 1.274 10⁴ N

This is the braking force and also the forces that couple the cars.

Answer:

5.4 m/s

Explanation:

Given that

Mass of the safe, m1 = 1000 kg

Distance to lower the safe, d = 3 m

Mass of furniture, m2 = 500 kg.

Speed of the safe, v = ?

To get the final speed by the time that the safe hits the truck, we first find its acceleration.

The total mass of the system is M = 1000 + 500 kg = 1500 kg

One of the forces acting on the system is that of gravity, and it acts on the safe friction acting on the furniture. Using the formula, we have

= m1*g - mu*m2g

= 1000 * 9.81 - 0.5 * 500 * 9.81

= 7357.5 N

From this calculated weight, we find the acceleration.

Acceleration, a = F/m

Acceleration, a = 7357.5 / 1500

Acceleration, a = 4.905 m/s²

From the question, we know that the Initial speed = 0 m/s

So, employing the use of one of the equations of motion, we have

v² - u² = 2aS

v² - 0 = 2 * 4.905 * 3

v² = 29.43

v = √29.43

v = 5.4 m/s

Answer:

b

Explanation:

The three places where fresh water can be found is ice sheets, water vapor, and rivers.

A fresh water is any type of naturally occurring liquid or frozen water containing low concentrations of dissolved salts and other total dissolved solids.

Fresh water can be found in the following locations;

Thus, the three places where fresh water can be found is ice sheets, water vapor, and rivers.

Learn more about fresh waters here: https://brainly.com/question/4432187

Answer:

[tex]1.5 \times 10^{-5} \mathrm{~T}[/tex].

Explanation:

Power carried by the line [tex]=P=450 \mathrm{MW}=450 \times 10^{6} \mathrm{~W}[/tex]

Voltage across the line Volts

Current flowing in the line =i

Size of magnetic field =B

Distance from the line

Formula Used:

Current flowing is given as

[tex]i=\frac{P}{\Delta V}[/tex]

Magnetic field by the current carrying wire is given as

[tex]B=\left(\frac{\mu}{4 \pi}\right)\left(\frac{2 i}{r}\right)[/tex]

Inserting the values

 [tex]B=\left(10^{-7}\right)\left(\frac{2(1500)}{(20)}\right) \\ B=1.5 \times 10^{-5} \mathrm{~T}[/tex]

Conclusion:

Thus, the magnetic field comes out to be [tex]1.5 \times 10^{-5} \mathrm{~T}[/tex].

Answer:

500 m/s

Explanation:

From the question,

V = √(T/m')................. Equation 1

Where V = Speed with which the wave travels on the string, T = Tension of the string, m' = Mass per unit length of the string.

Given: T = 1200 N, m' = 4.80×10⁻³ kg/m

Substitute these values into equation 1

V = √(1200/4.80×10⁻³)

V = √(250000)

V = 500 m/s

Hence the speed of the wave in the string is 500 m/s

Explanation:

c willl fall fast then a and b

Explanation:

Its D. The warm air from the land moves towards the water

One of the major effects of heat transfer is temperature change: heating increases the temperature while cooling decreases it. We assume that there is no phase change and that no work is done on or by the system. Experiments show that the transferred heat depends on three factors—the change in temperature, the mass of the system, and the substance and phase of the substance.

Figure a shows a copper-colored cylinder of mass m and temperature change delta T. The heat Q, shown as a wavy rightward horizontal arrow, is transferred to the cylinder from the left. To the right of this image is a similar image, except that the heat transferred Q prime is twice the heat Q. The temperature change of this second cylinder, which is also labeled m, is two delta T. This cylinder is surrounded by small black wavy lines radiating outward. Figure b shows the same two cylinders as in Figure a. The left cylinder is labeled m and delta T and has a wavy heat arrow pointing at it from the left that is labeled Q. The right cylinder is labeled two m and delta T and has a wavy heat arrow pointing to it from the left labeled Q prime equals two Q. Figure c shows the same copper cylinder of mass m and with temperature change delta T, with heat Q being transferred to it. To the right of this cylinder, Q prime equals ten point eight times Q is being transferred to another cylinder filled with water whose mass and change in temperature are the same as that of the copper cylinder.

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

[tex]m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss}[/tex] (1)

From the bottom to the top of the second hill

[tex]\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}[/tex] (2)

Where:

[tex]m[/tex] - Mass of the roller coaster car, in kilograms.

[tex]v_{1}[/tex] - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

[tex]g[/tex] - Gravitational acceleration, in meters per square second.

[tex]h_{1}[/tex] - Height of the first top of the hill with respect to the bottom, in meters.

[tex]W_{1, loss}[/tex] - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

[tex]v_{2}[/tex] - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

[tex]h_{2}[/tex] - Height of the second top of the hill with respect to the bottom, in meters.

[tex]W_{2, loss}[/tex] - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

[tex]m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}[/tex] (3)

Where [tex]W_{loss}[/tex] is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that [tex]m = 175\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]h_{1} = 18\,m[/tex], [tex]h_{2} = 8\,m[/tex] and [tex]v_{2} = 11\,\frac{m}{s}[/tex], then the work done by non-conservative force is:

[tex]W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right][/tex]

[tex]W_{loss} = 6574.75\,J[/tex]

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Answer:

0.15 s

Explanation:

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

Speed of sound (v) = 330 m/s

Distance (x) = 25 m

Time (t) =?

The time taken for the echo of the sound to the bat can be obtained as follow:

v = 2x / t

330 = 2 × 25 / t

330 = 50 / t

Cross multiply

330 × t = 50

Divide both side by 330

t = 50 / 330

t = 0.15 s

Thus, it will take 0.15 s for the echo of the sound to the bat

Answer:

13/23 ydfffgggggggffttf

Answer:

72 J

Explanation:

Use the Work formula

W= F x d

Given:

F - 48 N

d - 1.5 m

Solution:

W= F x d

W=  48 N x 1.5 m

W= 72 J

Answer:

  the maximum is I₁ axis of rotation at the end

     the minimum moment is I₂ axis of rotation at the center of mass

Explanation:

For this exercise we use the definition moment of inertia

          I = ∫ r² dm

for bodies of high symmetry it is tabulated; In this case we can approximate a broomstick to a thin rod, the moment of inertia with respect to a perpendicular axis when varying are

at one end

           I₁ = ⅓ mL²

in in center

           I₂ = [tex]\frac{1}{12}[/tex] m L²

There is another possible axis of rotation around the axis of the broom, in this case we have a solid cylinder

           I₃ = [tex]\frac{1}{2}[/tex] m r²

remember that the diameter of the broom is much smaller than its length, therefore this moment of inertia is very small

when examining the different moments of inertia:

     the maximum is I₁ axis of rotation at the end

     the minimum moment is I₂ axis of rotation at the center of mass