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pantera1 [17]
3 years ago
10

Which factors are used to calculate the kinetic energy of an object? Check all that apply.

Physics
1 answer:
Verizon [17]3 years ago
8 0
Gravity and velocity
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3 years ago
Help with questions please
solmaris [256]

Q9. R₁ only.

Since the upper branch is cut at x, the circuit in this branch is not closed. No current will flow through this branch. In this case, the circuit becomes a series circuit and current only flows through R₁.

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3 different resistors are connected in parallel, which means there will be three branches of parallel circuit, each with one resistor. Current can flow through 3 different branches, until the parallel circuits combine together again.

4 0
3 years ago
During a 40-mile trip, Marla traveled at an average speed of x miles per hour for the first y miles of the trip and at an averag
Ilya [14]

Answer:

\dfrac{32+0.2y}{40}

Explanation:

Total =40 miles.

y miles at x miles/hr.

40- y miles at 1.25 x miles/hr.\

We know that

Distance = time x velocity

For y miles

Lets time taken to cover y miles is t

y = x t

t=y/x                -----------1

For 40- y

Lets time taken to cover 40- y is t'

40- y = 1.25 x t'  

t'=(40-y)/1.25x           -----------2

By adding both equation

t+t'=y/x  + (40-y)/1.25x

t+t'=(32+0.2y)/x

Now lets time T taken by Marla if she will travel at x miles per hr for entire trip.

40 = x T

T=40/x

So

\dfrac{t+t'}{T}=\dfrac{32+0.2y}{40}

To find numerical value of above expression we have to know the value of y.

8 0
4 years ago
Josh starts his sled at the top of a 2.9-m-high hill that has a constant slope of 25∘. After reaching the bottom, he slides acro
algol13

Answer:

S=48.29 m

Explanation:

Given that the height of the hill h = 2.9 m

Coefficient of kinetic friction between his sled and the snow μ = 0.08

Let u be the speed of the skier at the bottom of the hill.

By applying conservation of energy at the top and bottom of the inclined plane we get.

Potential Energy=kinetic Energy

mgh = (1/2) mu²

u² = 2gh

u²=2(9.81)(2.9)

   =56.89

u=7.54 m/s

a = - f / m

a = - μ*m*g / m

a = - μg

From equation of motion

v²- u² = 2 -μ g S

v=0 m/s

-(7.54)²=-2(0.06)(9.81)S

S=48.29 m

3 0
3 years ago
A 7.77 kg mass is moving due west at 7.77 m/s. A second mass of 8.88 kg is moving due south at 8.88 m/s. What is the magnitude o
masya89 [10]

Answer:

8.362m/s

Explanation:

Given data

Mass m1= 7.77kg

Velocity v1= 7.77m/s

Mass m2= 8.88kg

Velocity v2= 8.88m/s

Apply the law of conservation of momentum for inelastic collision we have

m1v1+m2v2= (m+m2)V

7.77*7.77+ 8.88*8.88= (7.77+8.88)V

60.3729+78.8544= 16.65V

139.2273= 16.65V

Divide both sides by 16.65

V= 139.2273/16.65

V= 8.362m/s

Hence the final velocity is 8.362m/s

7 0
3 years ago
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