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mamaluj [8]
2 years ago
10

A toy car moves 8 min 4 s at the constant velocity. What is the car's velocity?

Physics
1 answer:
Schach [20]2 years ago
6 0

Explanation:

<u>Formula:</u>

velocity = (d \div t)

<u>d = distance given</u>

<u>t</u><u> </u><u>=</u><u> </u><u>the amount of time </u><u>given</u>

<u>Substitute the given values into the formula for velocity</u><u>:</u>

v = 8 \div 4

velocity is shortened for v.

8 (distance) divided by 4 (time) equals the velocity.

<u>Solve:</u>

2 = 8 \div 4

The velocity of the toy car equals: B. 2 m/s.

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kinetic energy

Explanation:

a certain amount of energy is transferred by the kick. The ball gains an equal amount of energy, mostly in the form of kinetic energy.

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What magnetically happens when there is an electric current in a wire?​
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2 years ago
Mary and her younger brother Alex decide to ride the 26 ft diameter carousel at the State Fair. Mary sits on one of the horses i
hammer [34]

Answer:

a_M=1.92a_A

Explanation:

\omega_M=\omega_A = Angular speed

r_M = Distance of Mary = 11.5 ft

r_A = Distance of Alex = 6 ft

Ratio of centripetal acceleration is given by

\dfrac{a_M}{a_A}=\dfrac{\omega_M^2r_M}{\omega_A^2r_A}\\\Rightarrow \dfrac{a_M}{a_A}=\dfrac{r_M}{r_A}\\\Rightarrow a_M=a_A\dfrac{r_M}{r_A}\\\Rightarrow a_M=\dfrac{11.5}{6}a_A\\\Rightarrow a_M=1.92a_A

Mary's centripetal acceleration is 1.92 times the centripetal acceleration of Alex

8 0
3 years ago
A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The
EleoNora [17]

Answer:

The tank is losing 4.976*10^{-4}  m^3/s

v_g = 19.81 \ m/s

Explanation:

According to the Bernoulli’s equation:

P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 +  \frac{1}{2}  \rho v_2^2 + \rho gh_2

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume  v_1 ≅ 0 ;

then v_2 can be determined as:\sqrt{[2g (h_1- h_2)]

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

v_2 = \sqrt{[2*9.81*(20 - 15)]

v_2 = \sqrt{[2*9.81*(5)]

v_2= 9.9 \ m/s  as it leaves the hole at the base.

radius r = d/2  = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = A_1v_2

J = πr²v_2    

J =\pi *(2*10^{-3})^{2}*9.9

J =1.244*10^{-4}  m^3/s

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : v_g = \sqrt{392.31}

v_g = 19.81 \ m/s

4 0
3 years ago
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