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3 years ago

A 2.10 X 10^3 kg car starts from rest at the top of a 5.0

Physics

1 answer:

xeze [42]3 years ago

4 0

Answer:

3.80 m/s

Explanation:

Solution: From the question we have:

A 2.10 x 10^3 kg car starts from rest at the top of a 5.0 m long driveway that is sloped at 20 degrees with the horizontal.

average friction force of 4.0 x 10^3 N impedes the motion

To find out: find the speed of the car at the bottom of the driveway.

=>h = 5×sin20 = 1.71 m.

PE = mgh = 2100×9.8×1.71 = 35,192 J.

KE = PE-work done by frictional force

KE=PE-4000×5

0.5mV^2 = 35192 - 4000×5

1050V^2 = 15,192

V^2 = 14.47

V = 3.80 m/s.

You might be interested in

You can describe the _______________ of an object by saying it is moving in a straight line or is curved around another object.

velikii [3]

Answer:

You can describe the motion of an object by saying it is moving in a straight line or is curved around another object. You can also describe where an object is by its position in relation to another object. The second object acts as a reference point. When an object changes position, you know it has motion. Motion can also be described by finding an object's speed or how fast or slow it moves in a certain amount of time. In addition, you can describe the object's speed AND direction together. This is called velocity

Explanation:

In the given answer-

Motion is defined as - the change in the movement or position of any object or body.

Position is said to be a place or somewhere or a location where any object or body is particularly placed/located or put on.

Reference point is a fixed point with regards to which any object or body changes its position. It is also called reference origin.

Speed is defined as the rate of any object covering certain distances. It is a scaler quantity (quantity which depends upon only magnitude).

Velocity is defined as the rate of speed per unit time. It is a vector quantity (quantity depending upon both magnitude and direction ).

7 0

3 years ago

it is about 384,750 kilometers from earth to the moon. it took the apollo astronauts about 2 days and 19.5 hours to fly to the m

Jet001 [13]

We know, speed = Distance / Time
d = 384,750 Km
t = 2 days, 19.5 hours = 48+19.5 = 67.5 hour

Substitute their values,
s = 384,750 / 67.5
s = 5700 Km/h

In short, Your Answer would be 5700 Km/h

Hope this helps!

7 0

3 years ago

Read 2 more answers

Which situation describes the highest rate of power?

Darya [45]

Answer : The correct option is, (D) A machine does 400 joules of work in 5 seconds.

Explanation :

Power : It is defined a the rate of doing work per unit time.

Formula used :

$P=\frac{w}{t}$

where,

P = power

w = work done

t = time

Now we have to determine the rate of power for the following options.

(A) A machine does 200 joules of work in 10 seconds.

$P=\frac{200}{10}=20W$

(B) A machine does 400 joules of work in 10 seconds.

$P=\frac{400}{10}=40W$

(C) A machine does 200 joules of work in 5 seconds.

$P=\frac{200}{5}=40W$

(D) A machine does 400 joules of work in 5 seconds.

$P=\frac{400}{5}=80W$

From this we conclude that, a machine does 400 joules of work in 5 seconds has the highest rate of power.

Hence, the correct option is, (D)

8 0

3 years ago

which planet should punch travel to if his goal is to weigh in at 118 lb? refer to the table of planetary masses and radii given

Harrizon [31]

The planet that Punch should travel to in order to weigh 118 lb is Pentune.

<h3 /><h3 /><h3>The given parameters:</h3>

Weight of Punch on Earth = 236 lb
Desired weight = 118 lb

The mass of Punch will be constant in every planet;

$W = mg\\\\m = \frac{W}{g}\\\\m = \frac{236}{g}$

The acceleration due to gravity of each planet with respect to Earth is calculated by using the following relationship;

$F = mg = \frac{GmM}{R^2} \\\\g = \frac{GM}{R^2}$

where;

M is the mass of Earth = 5.972 x 10²⁴ kg
R is the Radius of Earth = 6,371 km

For Planet Tehar;

$g_T =\frac{G \times 2.1M}{(0.8R)^2} \\\\g_T = 3.28(\frac{GM}{R^2} )\\\\g_T = 3.28 g$

For planet Loput:

$g_L =\frac{G \times 5.6M}{(1.7R)^2} \\\\g_L = 1.94(\frac{GM}{R^2} )\\\\g_L = 1.94g$

For planet Cremury:

$g_C =\frac{G \times 0.36M}{(0.3R)^2} \\\\g_C = 4(\frac{GM}{R^2} )\\\\g_C = 4 g$

For Planet Suven:

$g_s =\frac{G \times 12M}{(2.8R)^2} \\\\g_s = 1.53(\frac{GM}{R^2} )\\\\g_s = 1.53 g$

For Planet Pentune;

$g_P =\frac{G \times 8.3 }{(4.1R)^2} \\\\g_P = 0.5(\frac{GM}{R^2} )\\\\g_P = 0.5 g$

For Planet Rams;

$g_R =\frac{G \times 9.3M}{(4R)^2} \\\\g_R = 0.58(\frac{GM}{R^2} )\\\\g_R = 0.58 g$

The weight Punch on Each Planet at a constant mass is calculated as follows;

$W = mg\\\\W_T = mg_T\\\\W_T = \frac{236}{g} \times 3.28g = 774.08 \ lb\\\\W_L = \frac{236}{g} \times 1.94g =457.84 \ lb\\\\ W_C = \frac{236}{g}\times 4g = 944 \ lb \\\\ W_S = \frac{236}{g} \times 1.53g = 361.08 \ lb\\\\W_P = \frac{236}{g} \times 0.5 g = 118 \ lb\\\\W_R = \frac{236}{g} \times 0.58 g = 136.88 \ lb$

Thus, the planet that Punch should travel to in order to weigh 118 lb is Pentune.

The complete question is below:

Which planet should Punch travel to if his goal is to weigh in at 118 lb? Refer to the table of planetary masses and radii given to determine your answer.

Punch Taut is a down-on-his-luck heavyweight boxer. One day, he steps on the bathroom scale and "weighs in" at 236 lb. Unhappy with his recent bouts, Punch decides to go to a different planet where he would weigh in at 118 lb so that he can compete with the bantamweights who are not allowed to exceed 118 lb. His plan is to travel to Xobing, a newly discovered star with a planetary system. Here is a table listing the planets in that system (find the image attached).

In the table, the mass and the radius of each planet are given in terms of the corresponding properties of the earth. For instance, Tehar has a mass equal to 2.1 earth masses and a radius equal to 0.80 earth radii.

Learn more about effect of gravity on weight here: brainly.com/question/3908593

5 0

2 years ago

A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$

$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma *R^2$ since $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

$\mathbf{Q = 4 \pi R^2 \sigma }$

6 0

3 years ago