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3 years ago

Which of the following is equal to the area under a velocity-time graph

2 answers:

5 0

-- The area under a velocity/time graph, between two points in time, is the difference in displacement during that period of time.

-- The area under a speed/time graph, between two points in time, is the distance covered during that period of time.

Mashcka [7]3 years ago

4 0

Answer: Displacement

Explanation:

Apex

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A uniform soda can of mass 0.140 kg is 12.0 cm tall and filled with 0.354 kg of soda (Fig. 9-41). Then small holes are drilled i

Ugo [173]

Answer:

Explanation:

a ) The centre of mass of a can filled completely with any liquid of any mass lies at the center of the can or at height h /2 where h is height of the can . So centre of mass of can of mass .140 kg of height 12 cm will lie at height 6 cm . Similarly center of mass of its content will also lie at height 6 cm that is center point of the content .

Hence overall center of mass of can along with its content will lie at height 6 cm initially .

b )

After the can loses all the soda , can becomes empty . Hence its center of mass will lie at its center again .

c ) Its center of mass remains unchanged as the soda drains out completely .

d ) When soda starts draining out of the can , its lower part becomes heavier than the upper part . Hence its center of mass moves downwards . Its height is lowered . As height of soda is lowered in the can , center of mass also is lowered . It reaches its lowest point when soda in the can reaches half the height or when the can is half filled .

After that center of mass again starts rising upwards until the can is completely empty .

3 0

2 years ago

A body is projected vertically upwards with a speed 100 m/s from the ground then distance it covers in its last second of its up

Valentin [98]

The distance travelled by the body on last second of its upwards journey is 4.935m

What is projectile motion?

A projectile is an object or particle that is launched toward the surface of the Earth and moves along a curved path only under the influence of gravity.

To solve this question, consider the horizontal distance to be H and

distance traveled by the body in the last second be d

$H=U^{2} /2g$

$H=100^{2} /2*9.8$

H=510.2m

Now time taken to reach a height of

510.2 m= $\sqrt[n]{2H/g}$

=10.2 seconds

Now,

$s=ut+1/2gt^{2}$

$H-d= (100)(t-1)-1/2g(t-1)^{2}$

$510.2-d=100*9.2-1/2*9.8*9.2^{2}$

d=4936m

So, the correct answer is '4.936m'.

To know more about projectile motion visit:

brainly.com/question/11049671

#SPJ4

8 0

1 year ago

List important agents,or causes of erosion

Reil [10]

Erosion is caused naturally by water, wind or gravity, and accelerated erosion is caused by human use of land.

7 0

3 years ago

Read 2 more answers

Regardless of their frequency, wavelength, or energy, all electromagnetic waves: A. travel only through the vacuum of space. B.

agasfer [191]

All electromagnetic waves travel at the same speed in a vacuum: 3.0 x 10^5 (300,000) kilometres per second. some electromagnetic waves are part of the visible light spectrum and some do emit harmful radiation, but certainly not all. they travel fine on earth without the vacuum of space too.

6 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