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lapo4ka [179]
3 years ago
8

If 50 mL of each of the liquids in the answer choices were poured into a 250 mL beaker, which layer would be directly above a sm

all rubber ball with a density of 0.960 g/mL? A. sea water – density of 1.024 g/mL B. mineral oil – density of 0.910 g/mL C. distilled water – density of 1.0 g/mL D. petroleum oil – density of 0.820 g/mL
Physics
1 answer:
shutvik [7]3 years ago
6 0

Answer:

B. mineral oil – density of 0.910 g/mL.

Explanation:

Hello,

In this case, since the density is known as the degree of compactness a body has (mass in the occupied volume), the higher the density, the higher the weight of the body, therefore, if submerged into a liquid it could float if less dense than the liquid or sink if more dense than the liquid.

In such a way, since the rubber is more dense than mineral (0.960 g/mL > 0.910 g/mL) oil but less dense than distilled water (0.960 g/mL < 1.0 g/mL) we can say that B. mineral oil – density of 0.910 g/mL is directly above it when submerged.

Best regards.

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A 40-kg rock is dropped from an elevation of 10 m. What is the velocity of the rock when it is 5-m from the ground?
ivolga24 [154]

Answer:

Explanation:

Given

mass of rock m=40\ kg

Elevation of Rock h=10\ m

Distance traveled by rock with time

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

where, u=initial velocity

t=time

a=acceleration

here initial velocity is zero

when rock is 5 m from ground then it has traveled a distance of 5 m from top because total height is 10 m

5=0\times t+\frac{1}{2}(9.8)(t^2)

t^2=\frac{10}{9.8}

t=1.004\approx 1\ s

velocity at this time

v=u+at

v=0+9.8\times 1.004

v=9.83\ m/s

6 0
3 years ago
A plane travels 1743 KM in 2 hours 30 minutes. How fast was the plane traveling?
Advocard [28]

Answer:

v=697.2km/h

Explanation:

Hello.

In this case, since the velocity is computed via the division of the distance traveled by the elapsed time:

V=\frac{d}{t}

The distance is clearly 1743 km and the time is:

t=2h+30min*\frac{1h}{60min} =2.5h

Thus, the velocity turns out:

v=\frac{1743km}{2.5h}\\ \\v=697.2km/h

Which is a typical velocity for a plane to allow it be stable when flying.

Best regards.

5 0
3 years ago
. A 40.0-kg child standing on a frozen pond throws a 0.500-kg stone to the east with a speed of 5.00 m/s. Neglecting friction be
tamaranim1 [39]

Answer:

Explanation:

A 40kg child throw stone of 0.5kg

At a direction of 5m/s

Recoil can be calculated using recoil of a gun formula

m_1•v_1 + m_2•v_2

m_1•v_1 = -m_2•v_2

The negative sign show that the momentum of the boy is directed oppositely to that of the stone

m_1 Is mass of boy

v_1 is the recoil velocity of the boy

m_2 is mass of stone

v_2 is the velocity of stone

Then,

m_1•v_1 = -m_2•v_2

40•v_1 = -0.5 × 5

40•v_1 = -2.5

v_1 = -2.5 / 40

v_1 = -0.0625 m/s

The recoil velocity of the boy is 0.0625 m/s

6 0
3 years ago
A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal
AysviL [449]

Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

v=\sqrt{v_x^2+v_y^2}           (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

v_y^2=v_{oy}^2+2gh   (2)

vy: final velocity

voy: initial vertilal velocity = 0m/s  (because it is a semi parabolic motion)

g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}

vx is calculated by using the information about the horizontal range of the ball:

R=v_o\sqrt{\frac{2h}{g}}    (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}

The velocity of the ball just before it touches the ground is 46.99 m/s

5 0
3 years ago
1. Identify Which of the following is not a factor of global climate change:
Agata [3.3K]

Answer:

B.movement of continents

3 0
2 years ago
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