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3 years ago

13

Joaquin is at rest at the top of a hill on a skateboard. Four seconds later, he reaches the bottom of the hill at a final veloci

ty of 20 m/s. What was his acceleration?

1 answer:

Andre45 [30]3 years ago

8 0

Explanation:

Given: Initial Velocity=0

Final Velocity= 20 m/s

Time= 4 s

Accleration: Unknowns

Use kinematic equation

$v _{f} = v _{i} + at$

$20 = 0 + at$

$\frac{20m}{s} = a(4s)$

$a = \frac{5m}{ {s}^{2} }$

so a = 5 m/s^2

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Water waves approach an underwater "shelf" where the velocity changes from 2.8 m/s to 2.1 m/s. If the incident wave crests make

otez555 [7]

Answer:

24°

Explanation:

sin(34°)/sin(x)=v2/v1

x=arcsin(2,1*sin(34°)/2,8)=24°

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3 years ago

A robot is exploring charon, the dwarf planet pluto's largest moon. Gravity on charon is 0.278 meters per second [m divided b

DochEvi [55]

In order to find the efficiency first we will find the Change in Potential energy of the small stone that robot picked up

First we will find the mass of the stone

As it is given that stone is spherical in shape so first we will find its volume

$V = \frac{4}{3}\pi r^3$

$V = \frac{4}{3}\pi *(\frac{0.06}{2})^3$

$V = 1.13 * 10^{-4} m^3$

Now it is given that it's specific gravity is 10.8

So density of rock is

$\rho = 10.8 * 10^3 kg/m^3$

mass of the stone will be

$m = \rho V$

$m = 10.8* 10^3 * 1.13 * 10^{-4}$

$m = 1.22 kg$

now change in potential energy is given as

$\Delta U = mgH$

here

g = gravity on planet = 0.278 m/s^2

H = height lifted upwards = 15 cm

$\Delta U = 1.22* 0.278 * 0.15$

$\Delta U = 0.051 J$

Now energy supplied by internal circuit of robot is given by

$E = Vit$

V = voltage supplied = 10 V

i = current = 1.83 mA

t = time = 12 s

$E = 10* 1.83 * 10^{-3} * 12$

$E = 0.22 J$

Now efficiency is defined as the ratio of output work with given amount of energy used

$\eta = \frac{\Delta U}{E}*100$

$\eta = \frac{0.051}{0.22} = 0.23$

so efficiency will be 23 %

5 0

3 years ago

The base of a pyramid covers an area of 13.0 acres (1 acre = 43,560 ft2) and has a height of 481 ft. If the volume of a pyramid

Allisa [31]

V = 1/3 Bh v = 1/3 (13 ac)(43560ft^2/ac)(481ft) v = 90793560 ft^3 * 0.3048m/ft * 0.3048m/ft * 0.3048m/ft = 2570987m^3

3 0

3 years ago

Read 2 more answers

Please help I literally don’t understand

Veseljchak [2.6K]

Answer:

A= 2

B=3

C=4

D=5

E=7

F=8

H=12

7 0

3 years ago

A cylinder of mass 14.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed o

aev [14]

Answer:

a) 567J

b) 283.5J

c)850.5J

Explanation:

The expression for the translational kinetic energy is,

$E_r = \frac{1}{2} mv^2$

Substitute,

14kg for m

9m/s for v

$E_r = \frac{1}{2} (14) (9)^2\\= 567J$

The translational kinetic energy of the center of mass is 567J

(B)

The expression for the rotational kinetic energy is,

$E_R = \frac{1}{2} Iw^2$

The expression for the moment of inertia of the cylinder is,

$I = \frac{1}{2} mr^2$

The expression for angular velocity is,

$w = \frac{v}{r}$

substitute

1/2mr² for I

and vr for w

in equation for rotational kinetic energy as follows:

$E_R = (\frac{1}{2}) (\frac{1}{2} mr^2)(\frac{v}{r} )^2$

$= \frac{mv^2}{4}$

$E_R = \frac{14 \times 9^2 }{4} \\\\= 283.5J$

The rotational kinetic energy of the center of mass is 283.5J

(c)

The expression for the total energy is,

$E = E_r + E_R\\\\$

substitute 567J for E(r) and 283.5J for E(R)

$E = 567J + 283.5\\= 850.5J$

The total energy of the cylinder is 850.5J

6 0

3 years ago

Read 2 more answers