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2 years ago

Basketball is my favorite sport!

Physics

2 answers:

WARRIOR [948]2 years ago

6 0

Center.........................

natka813 [3]2 years ago

3 0

Answer:

the power forward(at least I think so)

Explanation:

volleyball is my favorite sport :D

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A 1560-kilogram truck moving with a speed of 28.0 m/s runs into the rear end of a 1070-kilogram stationary car. If the collision

Nimfa-mama [501]

Answer:

Δ KE = 249158.6 kJ

Explanation:

given data

Truck mass M = 1560 Kg

Truck initial speed, u = 28 m/s

mass of car m = 1070 Kg

initial speed of car u1 = 0 m/s

solution

first we get here final speed by using conservation of momentum that is express as

Mu = (M+m) V .......................1

put here value we get

1560 × 28 = (1560 + 1070 ) V

solve it we get

final speed V = 16.60 m/s

and

Change in kinetic energy will be here

Δ KE = $\frac{1}{2} Mu^2 - \frac{1}{2}(M+m)V^2$ .................2

put here value and we get

Δ KE = $\frac{1}{2}\times 1560\times 28^2 - \frac{1}{2}\times (1560 + 1070)\times 16.60^2$

solve it we get

Δ KE = 249158.6 kJ

6 0

2 years ago

An insulating cup contains 200 grams of water at 25 ∘C. Some ice cubes at 0 ∘C is placed in the water. The system comes to equil

Nataly [62]

Answer:

The amount of ice added in gram is 32.77g

Explanation:

This problem bothers on the heat capacity of materials

Given data

Mass of water Mw= 200g

Temperature of water θw= 25°c

Temperature of ice θice= 0°c

Equilibrium Temperature θe= 12°c

Mass of ice Mi=???

The specific heat of ice Ci= 2090 J/(kg ∘C)

specific heat of water Cw = 4186 J/(kg ∘C)

latent heat of the ice to water transition Li= 3.33 x10^5 J/kg

heat heat loss by water = heat gained by ice

N/B let us understand something, heat gained by ice is in two phases

Heat require to melt ice at 0°C to water at 0°C

And the heat required to take water from 0°C to equilibrium temperature

Hence

MwCwΔθ=MiLi +MiCiΔθ

Substituting our data we have

200*4186*(25-12)=Mi*3.3x10^5+

Mi*2090(12-0)

837200*13=Mi*3.3x10^5+Mi*2090

10883600=332090Mi

Mi=10883600/332090

Mi= 32.77g

4 0

3 years ago

What is the difference between speed and volocity?

Sliva [168]

The average speed is the distance per time ratio. velocity is the rate at which the position changes

3 0

2 years ago

The period T of a pendulum of length L is measured to determine g at the surface of Earth. The equation used is T=2π√L/g. The ma

saul85 [17]

Answer:

C: Variation in the value of g as the pendulum bob moves along its arc.

Explanation:

The formula for period of a simple pendulum is given by;

T = 2π√(L/g)

Where;

L is length

g is acceleration due to gravity

Now, from this period equation, it is clear that the only thing that can affect the period of a simple pendulum are changes to its length and acceleration due to gravity.

Looking at the options, the only one that talks about either the length or gravity as being potential causes of the error is option C

4 0

2 years ago

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is W/m². What is the rms

inn [45]

I think your question should be:

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is

$S = 1.23*10^9 W/m^2$

What is the rms value of (a) the electric field and

(b) the magnetic field in the electromagnetic wave emitted by the laser

Answer:

a) $6.81*10^5 N/c$

b) $2.27*10^3 T$

Explanation:

To find the RMS value of the electric field, let's use the formula:

$E_r_m_s = sqrt*(S / CE_o)$

Where

$C = 3.00 * 10^-^8 m/s$ ;

$E_o = 8.85*10^-^1^2 C^2/N.m^2$ ;

$S = 1.23*10^9 W/m^2$

Therefore

$E_r_m_s = sqrt*{(1.239*10^9W/m^2) / [(3.00*10^8m/s)*(8.85*10^-^1^2C^2/N.m^2)]}$

$E_r_m_s= 6.81 *10^5N/c$

b) to find the magnetic field in the electromagnetic wave emitted by the laser we use:

$B_r_m_s = E_r_m_s / C$ ;

$= 6.81*10^5 N/c / 3*10^8m/s$ ;

$B_r_m_s = 2.27*10^3 T$

8 0

3 years ago

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