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Free_Kalibri [48]
3 years ago
8

Plz help I don’t understand...

Physics
1 answer:
marysya [2.9K]3 years ago
4 0
Message me back and I will help :)
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Test your prediction through calculation for the situations of the clay bob and the bouncy ball. Assume each has a mass of 100 g
melamori03 [73]

Answer:

a) Δp = -2.0 kgm / s,  b)   Δp = -4 kg m / s

Explanation:

In this exercise the change in moment of a ball is asked in two different cases

a) clay ball, in this case the ball sticks to the door and we have an inelastic collision where the final velocity of the ball is zero

         Δp = p_f - p₀

         Δp = 0 - m v₀

         Δp = - 0.100 20

         Δp = -2.0 kgm / s

b) in this case we have a bouncing ball, this is an elastic collision, as the gate is fixed it can be considered an object of infinite mass, therefore the final speed of the ball has the same modulus of the initial velocity, but address would count

         v_f = - v₀

        Δp = p_f -p₀

        Δp = m v_f - m v₀

        Δp = m (v_f -v₀)

        Δp = 0.100 (-20 - 20)

        Δp = -4 kg m / s

6 0
3 years ago
How many calories are needed to raise the temperature of 2g of water 3 C?
olga nikolaevna [1]

Answer:

The number of calories needed is 6c.

Explanation:

The amount of energy Q needed to raise the temperature \Delta T of water of mass m is

Q = mC\Delta T

where C = 1cal/g\cdot C is the specific heat capacity of water.

Putting in numbers into equation (1), we get:

Q = (2g)(1)(3^oC )\\

\boxed{Q = 6\:cal }

which is the number of calories needed.

8 0
3 years ago
Exercice 1
Mariana [72]

Answer:

Injections of aqueous solution of fruct

levulose of formula C, H, O,,

For

prevent dehydration such solutions are obtained

dissolving a mass m = 25g of fructose for a volume of 50

final solution.

1.1 Calculate the molecular molar mass of fructose

1.2 Determine the amount of corresponding fructose material

1.3 Calculate the molar concentration of these fructose solutions

1.4

6 0
3 years ago
A proton, initially traveling in the +x-direction with a speed of 5.05×10^5 m/s , enters a uniform electric field directed verti
Korvikt [17]

Answer:

The strength of the electric field is 1.35\times10^{4}\ N/C.

Explanation:

Given that,

Speed v= 5.05\times10^{5}\ m/s

Time t= 3.90\times10^{-7}\ s

Angle = 45°

We need to calculate the acceleration

Using equation of motion

v = u+at

5.05\times10^{5}=0+a\times3.90\times10^{-7}

a =\dfrac{5.05\times10^{5}}{3.90\times10^{-7}}

a=1.29\times10^{12}\ m/s^2

We need to calculate the strength of the electric field

Using relation of newton's second law and electric force

F= ma=qE

ma = qE

E=\dfrac{ma}{q}

Put the value into the formula

E=\dfrac{1.67\times10^{-27}\times1.29\times10^{12}}{1.6\times10^{-19}}

E=1.35\times10^{4}\ N/C

Hence, The strength of the electric field is 1.35\times10^{4}\ N/C.

3 0
3 years ago
Pls help on this one?
sattari [20]
The answer is point C
4 0
3 years ago
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