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Zinaida [17]
3 years ago
15

A drag racer starts from the rest and accelerates at 7.4m/s^2.how far will he travel in 2.0 seconds?

Physics
1 answer:
Vesna [10]3 years ago
8 0

Answer:

14.8 m

Explanation:

S= ut + \frac{1}{2}at^{2}

where u = initial velocity

S= (0 \frac{m}{s})(2s) + \frac{1}{2}(7.4\frac{m}{s^{2} })(2s^{2})

S=  \frac{1}{2}(7.4\frac{m}{s^{2} })(2s^{2})

S=14.8 m

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What happens to momentum when two objects collide
Lostsunrise [7]

Answer:

The sum of momenta of each object before and after collision is equal . Thus when they collide , their momentum will change but its sum will be sane as the sum of the momentum before collision

3 0
3 years ago
Sophia is planning on going down an 8-m water slide. Her weight is 50 N. She knows that she has gravitational potential energy (
RideAnS [48]

Answer:

Explanation:

graph would be a straight line from (0, 0) to (400, 8)

Plot points are

PE = mgh

50(0) = 0 J

50(2) = 100 J

50(4) = 200 J

50(6) = 300 J

50(8) = 400 J

4 0
3 years ago
Which substance is a combination of different atoms?
Lady_Fox [76]

Answer:

The answer is compound

Explanation:

heterogeneous mixture is wronggggg

5 0
3 years ago
In certain cases, using both the momentum principle and energy principle to analyze a system is useful, as they each can reveal
kramer

Explanation:

The gravitational force equation is the following:

F_G = G * \frac{m_1 m_2}{r^2} \\

Where:

G = Gravitational constant = 6.67408 * 10^{-11} m^3 kg^{-1} s^{-2}

m1 & m2 = the mass of two related objects

r = distance between the two related objects

The problem gives you everything you need to plug into the formula, except for the gravitational constant. Let me know if you need further clarification.

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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