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A 10 kg block moving at 10 m/s in a direction 45 degrees above the horizontal. When it has fallen to a point that is 10 m below the initial point measured vertically (without air friction), the block's kinetic energy is closest to

Answers

Answer 1

Answer:

The block's kinetic energy is closest to 1500 Joules.

Kinetic energy :

The energy is always conserved.

So that, the total kinetic energy will be sum of initial potential energy and kinetic energy during falling.

Given that, mass(m)=10kg, v=10m/s, h=10m,g=10m/s^2

K.E=(1/2)mv^2 + mgh

K.E=(1/2)*10*100 + (10*10*10)

K.E=500 + 1000=1500Joule

The block's kinetic energy is closest to 1500 Joules.

Learn more about the kinetic energy here:

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Answer 2

Answer:

Answer:

Kinetic energy = 1500 J

Explanation:

The computation of the block's kinetic energy is shown below:

As we know that

Conservation of energy is

PE_i + KE_i = PE_f + KE_f

where,

Initial Potential energy = PE_i = m gh = 10kg× 10m/s^2 × 10m = 1000 J

Initial Kinetic energy = KE_i = (0.5) m V^2 = (0.5) (10 kg) (10 m/s)^2 = 500 J

Final potential energy = PE_f = mgh = 0

As h = 0 which is at reference line

PE_i + KE_i = PE_f + KE_f

Now put these valeus to the above formulas

1000 J + 500 J = 0 + KE_f

After solving this

Kinetic energy = 1500 J

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What form of braking is used to bring a vehicle to a smooth stop by applying smooth,steady pressure to the braketrail braking
controlled braking
threshold braking
coasting

Answers

Answer:

Controlled braking

Explanation:

CONTROLLED BRAKING occur in a situation where a person or an individual driving a vehicle releases the brake and slowly apply smooth as well as firmly pressure on the brake without the wheels been locked which is why CONTROLLED BRAKING are often used for emergency stops by drivers reason been that it helps to reduce speed when driving as fast as possible while the driver maintain the steering control of the vehicle.

Therefore the form of braking which is used to bring a vehicle to a smooth stop by applying smooth,steady pressure to the brake is called CONTROLLED BRAKING.

Final answer:

The method of braking that involves applying smooth, steady pressure to the brake to bring the vehicle to a smooth stop is called controlled braking. It helps prevent skidding and provides the driver with more control over the vehicle.

Explanation:

The form of braking used to bring a vehicle to a smooth stop by applying smooth, steady pressure to the brake is known as controlled braking. This method of braking involves applying consistent, even pressure to the brake pedal, which allows the car to slow down gently and gradually. It helps prevent uncontrolled skidding and provides the driver with more control over the vehicle's direction and speed during the stop. Unlike other methods like trail braking, threshold braking, or coasting, controlled braking is typically the safest and most effective method for daily driving conditions.

Learn more about Controlled Braking here:

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A pair of thin spherical shells with radius r and R, r < R are arranged to share a center. What is the capacitance of the system. If a potential difference V is created between the shells, how much energy is stored between them?

Answers

Answer:

Capacitance = ( 4π×∈×r×R ) / (R-r)

energy store = ( 4π×∈×r×R )×V² / (R-r)

Explanation:

given data

radius = r

radius = R

r < R

to find out

capacitance and how much energy store

solution

we consider here r is inner radius and R is outer radius

so now apply capacitance C formula that is

C = Q/V .................1

here Q is charge and V is voltage

we know capacitance have equal and opposite charge so

V = $\int\limits^R_r {E} \, dx$

here E = Q / 4π∈k²

V = Q / 4π∈ $\int\limits^R_r {1/k^2} \, dx$

V = Q / 4π∈ × ( 1/r - 1/R )

V = Q(R-r) / ( 4π×∈×r×R )

so from equation 1

C = Q/V

Capacitance = ( 4π×∈×r×R ) / (R-r)

and

energy store is 1/2×C×V²

energy store = ( 4π×∈×r×R )×V² / (R-r)

Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)

Answers

Answer:

1/4F

Explanation:

We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.

So F α Qq

But if it is now half the initial charges, then

F α (1/2)Q *(1/2)q

F α (1/4)Qq

Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.

Thus the answer will be 1/4F

The brakes of a car moving at 14m/s are applied, and the car comes to a stop in 4s. (a) What was the cars acceleration? (b) How long would the car take to come to a stop starting from 20m/s with the same acceleration? (c) How long would the car take to slow down from 20m/s to 10m/s with the same acceleration?

Answers

(1) The acceleration of the car will be $a=-3.5(m)/(s^2)$

(2) The time taken $t=5.7s$

(3) The time is taken by the car to slow down from 20m/s to 10m/s $t=2.9s$

What will be the acceleration and time of the car?

(1) The acceleration of the car will be calculated as

$a=(v-u)/(t)$

Here

u= 14 $(m)/(s)$

$a=(0-14)/(4) =-3.5(m)/(s^2)$

(2) The time is taken for the same acceleration to 20 $(m)/(s)$

$a=(v-u)/(t)$

$t=(v-u)/(a)$

u=20 $(m)/(s)$

$t=(0-20)/(-3.5) =5.7s$

(3) The time is taken to slow down from 20m/s to 10m/s with the same acceleration

From same formula

$t=(v-u)/(a)$

v=10 $(m)/(s)$

u=20 $(m)/(s)$

$t=(10-20)/(-3.5) =2.9s$

Thus

(1) The acceleration of the car will be $a=-3.5(m)/(s^2)$

(2) The time taken $t=5.7s$

(3) The time is taken by the car to slow down from 20m/s to 10m/s $t=2.9s$

To know more about the Equation of the motion follow

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(a) $-3.5 m/s^2$

The car's acceleration is given by

$a=(v-u)/(t)$

where

v = 0 is the final velocity

u = 14 m/s is the initial velocity

t = 4 s is the time elapsed

Substituting,

$a=(0-14)/(4)=-3.5 m/s^2$

where the negative sign means the car is slowing down.

(b) 5.7 s

We can use again the same equation

$a=(v-u)/(t)$

where in this case we have

$a=-3.5 m/s^2$ is again the acceleration of the car

v = 0 is the final velocity

u = 20 m/s is the initial velocity

Re-arranging the equation and solving for t, we find the time the car takes to come to a stop:

$t=(v-u)/(a)=(0-20)/(-3.5)=5.7 s$

(c) $2.9 s$

As before, we can use the equation

$a=(v-u)/(t)$

Here we have

$a=-3.5 m/s^2$ is again the acceleration of the car

v = 10 is the final velocity

u = 20 m/s is the initial velocity

Re-arranging the equation and solving for t, we find

$t=(v-u)/(a)=(10-20)/(-3.5)=2.9 s$

Tarzan, whose mass is 96 kg, is hanging at rest from a tree limb. Then he lets go and falls to the ground. Just before he lets go, his center of mass is at a height 2.1 m above the ground and the bottom of his dangling feet are at a height 1.3 above the ground. When he first hits the ground he has dropped a distance 1.3, so his center of mass is (2.1 - 1.3) above the ground. Then his knees bend and he ends up at rest in a crouched position with his center of mass a height 0.4 above the ground.Consider the point particle system. What is the speed v at the instant just before Tarzan's feet touch the ground?

Answers

Answer:

5.05 m/s

Explanation:

The distance from the bottom of his feet to his center of mass is (when is hanging at rest) is 2.1 - 1.3 = 0.8 m. Assume he keeps the posture, as soon as his feet touches the ground, his center of mass is 0.8 m above the ground. This would mean that he has traveled a distance of 2.1 - 0.8 = 1.3 m vertically. Using the law of energy conservation for potential and kinetic energy, also let the ground be ground 0 for potential energy, we have the following mechanical conservation energy:

$mgH = mgh + mv^2/2$

Since he was hanging at rest, his initial kinetic energy at H = 2.1m must be 0. Let g = 9.81m/s2 and m be his mass, we can calculate for his velocity v at h = 0.8 m. First start by dividing both sides by m

$gH = gh + v^2/2$

$v^2 = 2g(H - h)$

$v^2 = 2*9.81(2.1 - 0.8) = 25.506$

$v = √(25.506) = 5.05 m/s$

Analysis of the relationship between the fuel economy (mpg) and engine size (liters) for 35 models of cars produces the regression model ModifyingAbove mpg with caret equals 36.44 minus 3.829 times Engine size. If a car has a 5 liter engine, what does this model suggest the gas mileage would be?