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3 years ago

I need help please only way to put my grade up !!!!!! Would appreciate it !!! Someone who’s good at this

Physics

1 answer:

motikmotik3 years ago

3 0

Answer:

Total energy = 1000J

KE = 500J

PE = 500J

Explanation:

As you may know, the equation for gravitational potential energy is mgh (weight x height)

If the skateboard is halfway down, that means it is at half the height. As the skateboard speeds up (as it goes downward), the potential energy becomes kinetic energy. Since it has 500J of kinetic energy at half way down, it means it had double that amount of Potential energy at the top (1000J). Since half of that became kinetic energy, there is only 500J of PE left.

Total energy = KE + PE = 1000J

KE = 500J

PE = 500J

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Please help me out !!

blagie [28]

I think the answer would be letter a

4 0

2 years ago

What is the energy required to change 600g of liquid into steam if the specific heat latent is 4000 J/kg

Rzqust [24]

Answer:

2400 J

Explanation:

Latent heat: This is also called hidden heat, it is the heat that is not detectable by the thermometer.

From the question,

Q = cm.................. Equation 1

Where Q = Energy, c = specific latent heat of the liquid, m = mass of the liquid.

Given: c = 4000 J/kg, m = 600 g =( 600/1000) kg = 0.6 kg

Substitute these values into equation 1

Q = 4000×0.6

Q = 2400 J

Hence the energy required is 2400 J

6 0

3 years ago

Use the definition of scalar product, a with arrow · b with arrow = ab cos θ, and the fact that a with arrow · b with arrow = ax

Zolol [24]

Answer:

5+5=10

4+6=10

angle between is 45

3 0

3 years ago

A 7.80 g bullet moving at 530 m/s penetrates a tree trunk to a depth of 5.40 cm. (a) Use work and energy considerations to find

True [87]

Answer:

a) F=20287.22N

b) t=2*10^-4s

Explanation:

E=1/2*m*v^2=0.5*7.8*10^-3kg*(530m/s)^2=1095.51J

The frictional force's work must be equal than the energy to stop the bullet.

So: W=F*d=F*0.054m=1095.51J, F=20287.22N

Considering the frictional force is constant, the bullet moves with constant aceleration.

a=F/m=20287.22N/7.8*10-2kg=2.6*10^6m/s^2

then d(t)=Vt-1/2*a*t^2,

5.4*10^-2m=530m/s*t-1.3*10^6m/s^2*t^2

I will calculate the time using the cuadratic formula:

$\frac{-b+-\sqrt{b^{2}-4ac } }{2a}$

with a=1.3*10^6, b=-530, c=5.4*10^-2

t=2*10^-4s

8 0

3 years ago

Se lanza una pelota de hule elástica de masa mA con una velocidad VAi contra la puerta trasera de un camión muy pesado de masa m

miv72 [106K]

Answer:

$v_{A}=v_{Ai}$

Explanation:

The total momentum of the system must conserve by the law of momentum conservation.

$p_b=p_a\\\\m_Av_{Ai}+m_Bv_{Bi}=m_Av_{A}+m_Bv_{B}$ ( 1 )

If you consider the shock between the ball and the truck as totally elastic, and also if the truck is much more massive than the ball, you can assume that the truck does not move when the ball hits it.

Then the truck is always at rest

Hence, you have in the expression ( 1 ):

$m_Av_{Ai}+0=m_Av_{A}+0\\\\v_{Ai}=v_{A}$

3 0

3 years ago