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3 years ago

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An object's (mass or weight) will remain constant throughout the universe, but its (mass or weight) can change from planet to pl

anet. If you increase the mass of a planet, what happens to its gravity? (Strength of gravity increases, strength of gravity decreases, strength og gravity stops completely) If the gravity on a planet decreases, what happens to the weight of an object on that planet? (Weight increases, weight decreases, weight stays the same)

2 answers:

melisa1 [442]3 years ago

8 0

Answer:

1: mass

2:weight

3: gravity increases

4: it decreases

Hope this helps:)

Explanation:

3 years ago

4 Is weight decreases

krek1111 [17]3 years ago

6 0

Answer: 1. An objects mass

2. But it’s weight

3. Strength of gravity increases

4 weight decreases

Explanation:

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Cyclist A is moving at 20.0 m/s whereas cyclist B is moving at 12.0 m/s in the same direction and is initially ahead ofA. When t

stealth61 [152]

Answer:

$v_{fA} = 28 \frac{m}{s}$

Explanation:

The kinematic parameters from the moment the two cyclists begin to accelerate until they meet are:

Initial parameters:

$v_{oA} = 20 m/s$ : Initial speed of cyclist A

$v_{oB} = 12 m/s$ : Initial speed of cyclist B

Final parameters:

$v_{fB} = 20 m/s$ : Final speed of cyclist B

$d_{A} = d_{B}$

distance of cyclist A = distance of cyclist B

$t_{A} =t_{B} = 12s$

time of cyclist A = time of cyclist B

Cyclist B Kinematics

$v_{fB} = v_{oB} +a_{B} *t\\$

$36= 12 + a_{B} *12$

$a_{B} = \frac{36-12}{12}$

$a_{B} = 2 \frac{m}{s^{2} }$

$d_{B} =( v_{oB})*( t)+( \frac{1}{2} )*(a_{B})* (t)^{2}$

$d_{B} =( 12*(12)+( \frac{1}{2} )*(2)* (144)}$

$d_{B} = 288m$

Cyclist A Kinematics

$d_{A} =( v_{oA})*( t)+( \frac{1}{2} )*(a_{A})* (t)^{2}$

$d_{A} =(20*( 12)+( \frac{1}{2} )*(a_{A})* 144}$

$d_{A} = 240 + 72*(a_{A})$

$288=240+72*(a_{A} )$

$a_{A} = \frac{288-240}{72}$

$a_{A} = 0.67 \frac{m}{s^{2} }$

$v_{fA} = v_{oA} + a_{A} * t$

$v_{fA} = 20 + 0.67*12$

$v_{fA} = 28 \frac{m}{s}$

7 0

3 years ago

A simple harmonic oscillator of amplitude A has a total energy E.

jeka57 [31]

Answer:

a) $K=E-\frac{kA^2}{18}$

b) $U=\frac{kA^2}{18}$

Explanation:

The law of conservation of mechanical energy states that total mechanical energy remains constant during oscillation. Mechanical energy is defined as the sum of kinetic energy and potential energy:

$E=U+K\\E=\frac{kx^2}{2}+\frac{mv^2}{2}$

a) The position is one-third the amplitude. So, we have $x=\frac{1}{3}A$ . Replacing and solving for K

$E=\frac{k(\frac{1}{3}A)^2}{2}+K\\E=\frac{kA^2}{18}+K\\K=E-\frac{kA^2}{18}$

b) The potential energy is defined as:

$U=\frac{kx^2}{2}$

Replacing:

$U=\frac{k(\frac{1}{3}A)^2}{2}\\U=\frac{kA^2}{18}$

4 0

3 years ago

Find the magnitude & direction: 50N 25 N 35 N 10N

adoni [48]

F1x + F2x = Rx

↓

Rx = F1x + F2x

↓

Rx = F1 cos45° + F2

↓

Rx = (50N)(cos45°) + 60N

↓

Rx = 95N

Similarly, if we sum all the y components, we will get the y component of the resultant force:

F1y + F2y = Ry

↓

Ry = F1y + F2y

↓

Ry = F1 sin45° + 0

↓

Ry = F1 sin45°

↓

Ry = (50N)(sin45°)

↓

Ry = 35N

At this point, we know the x and y components of R, which we can use to find the magnitude and direction of R:

Rx = 95N

Ry = 35N

8 0

3 years ago

A spring with spring constant of 26 N/m is stretched 0.22 m from its equilibrium position. How much work must be done to stretch

Evgen [1.6K]

Answer:

1.503 J

Explanation:

Work done in stretching a spring = 1/2ke²

W = 1/2ke²........................... Equation 1

Where W = work done, k = spring constant, e = extension.

Given: k = 26 N/m, e = (0.22+0.12), = 0.34 m.

Substitute into equation 1

W = 1/2(26)(0.34²)

W = 13(0.1156)

W = 1.503 J.

Hence the work done to stretch it an additional 0.12 m = 1.503 J

8 0

3 years ago

A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend

Viefleur [7K]

a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how

long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175

s this is the time to fall from the top; it would take the same time to travel

upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175

= 0.35s

b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice

to solve this problem the time it takes to fall the final 0.13 m is: time it

takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to

fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it

takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m

is then twice this, or 0.08s

5 0

4 years ago

Read 2 more answers