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3 years ago

A car accelerates at 2 meters/s/s. Assuming the car starts from rest how far will it travel in 10 seconds

Physics

2 answers:

Kruka [31]3 years ago

8 0

Answer:

Distance = velocity x time, so 10 m/s X 10 s = 100 m

Explanation:

If you accelerate at 2 m/s^2 for 10 seconds, at the end of the 10 seconds you are moving at a rate of 20 m/s.

V(f) = V(i) + a*t

Final velocity = initial velocity + acceleration x time

Your average velocity will be half of your final, because you accelerated at a constant rate. So your average velocity is 10 m/s.

Distance = velocity x time, so 10 m/s X 10 s = 100 m

LiRa [457]3 years ago

4 0

Answer:

100 m

Explanation:

Given,

Initial velocity ( u ) = 0 m/s

Acceleration ( a ) = 2 m/s^2

Time ( t ) = 10 sec s

To find : Displacement ( s ) = ?

By 2nd equation of motion,

s = ut + at^2 / 2

= ( 0 ) ( 10 ) + ( 2 ) ( 10 )^2 / 2

= 0 + ( 2 ) ( 100 ) / 2

= 200 / 2

s = 100 m

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As the resistance increased under 25 V, the current blank. Compared to calculated currents, experimental currents proved to be b

sukhopar [10]

Answer:

Decreased, The Same, 2.5 and .25

Explanation:

4 0

3 years ago

A cannon is fired horizontally from atop a 40.0 m tower. The cannonball travels 175 m horizontally before it strikes the ground.

Anastasy [175]

The vertical movement of the projectile is described by:

y = H - gt² / 2

When the cannonball is on the ground, y= 0 so:

0 = H - gt² /2

Solving for t:

$t = \sqrt{\dfrac{2H}{g}} = \sqrt{\dfrac{2(40\;m)}{9.8\;m/s^2}} = 2.8\;s$

The horizontally movement of the projectile is described by:

x = v₀t

Solving for v₀:

v₀ = x/t

v₀ = 175 m / 2.8 s

v₀ = 61.5 m/s

3 0

3 years ago

A clam of mass 0.12 kg dropped by a seagull takes 3.0 s to hit the ground. [Neglect friction.]

inna [77]

This is a defective, misleading question, and should never be asked in a Physics class.

There is no such thing as the force due to the impact.

If you know how long it takes the clam to stop once it begins to hit the dirt,
then you can calculate the impulse transferred to it, and tease a force out
of that. But the question doesn't give us the time.

It depends on the material of the surface. Was the clam dropped onto dirt ?
Into a dumpster ? Onto grass ? Concrete ? Styrofoam ? Mud ? The answer
is different in each case, and we still need to know the short length of time
AFTER it first encountered whatever surface brought it to rest.

I would kick this question back to the Physics teacher. It's meaningless,
and the longer you try to work on it, the more nonsense you'll plant into
your head that'll need to be dug out later.

8 0

3 years ago

Several large firecrackers are inserted into the holes of a bowling ball, and the 6.3 kg ball is then launched into the air with

Anni [7]

Answer

given,

mass of the ball = 6.3 kg

speed of the ball = 10.4 m/s

angle made with horizontal = 43°

m_a = 1.8 kg v_a = 2.2 m/s

m_b = 1.6 kg v_b = 1.8 m/s

mass of third particle = 6.3 - 1.8 - 1.6

= 2.9 kg

u cos θ = 10.4 x cos 43° = 7.61 m/s

by using conservation momentum along x-axis

6.3 x 7.61 = 1.8 × (-2.2) + 0 + 1.6 × V₃ₓ

V₃ₓ = 32.44 m/s (toward right)

by using conservation momentum along y-axis

0 = 0 + 1.6 x 1.8 + 1.6 × V₃y

V₃y = -1.8 m/s (indicate downward)

velocity of the third particle

$v = \sqrt{32.44^2 + (-1.8)^2}$

v = 32.49 m/s

$tan \theta = \dfrac{-1.8}{32.44}$

$\theta = tan^{-1}(\dfrac{-1.8}{32.44})$

θ = 3.176° (downward with horizontal)

4 0

3 years ago

A 78.5-kg man floats in freshwater with 3.2% of his volume above water when his lungs are empty, and 4.85% of his volume above w

Dima020 [189]

Answer:

A) V_air = 1.295 L

B) Volume is not reasonable

Explanation:

A) Let;

m be total mass of the man

m_p be the mass of the man that pulled out of the water because of the buoyant force that pulled out of the lung

m_3 be the mass above the water with the empty lung

m_5 be the mass above the water with full lung

F_b be the buoyant force due to the air in the lung

V_a be the volume of air inside man's lungs

w_p be the weight that the buoyant force opposes as a result of the air.

Now, we are given;

m = 78.5 kg

m_3 = 3.2% × 78.5 = 2.512 kg

m_5 = 4.85% × 78.5 = 3.80725 kg

Now, m_p = m_5 - m_3

m_p = 3.80725 - 2.512

m_p = 1.29525 kg

From archimedes principle, we have the formula for buoyant force as;

F_b = (m_displaced water)g = (ρ_water × V_air × g)

Where ρ_water is density of water = 1000 kg/m³

Thus;

F_b = w_p = 1.29525 × 9.81

F_b = 12.7064 N

As earlier said,

F_b = (ρ_water × V_air × g)

Thus;

V_air = F_b/(ρ_water × × g)

V_air = 12.7064/(1000 × 9.81)

V_air = 1.295 × 10^(-3) m³

We want to convert to litres;

1 m³ = 1000 L

Thus;

V_air = 1.295 × 10^(-3) × 1000

V_air = 1.295 L

B) From research, the average lung capacity of an adult human being is 6 litres of air.

Thus, the calculated lung volume is not reasonable

4 0

3 years ago