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2 years ago

(c) A solid weights 64N in air and 48N when totally immersed in a liquid 0.8g/cm3.

1 answer:

8 0

(a) When immersed in the liquid, the solid displaces an amount of liquid weighing 64 N - 48 N = 16 N, and this is the same as the magnitude of the buoyant force/upthrust.

(b) Archimedes' principle says the solid has a volume equal to that of the displaced liquid. At a density of 0.8 g/cm³ = 0.0008 kg/cm³, a 16-N amount of this liquid has a volume of

(16 N) / g × (1/0.0008 cm³/kg) = 2040.82 cm³

(c) Density is mass per unit volume, so you get the solid's density by dividing its mass (weight/g) by the volume in (b):

((64 N) / g) / (2040.82 cm³) ≈ 0.0032 kg/cm³ ≈ 3.2 g/cm³

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klasskru [66]

Answer:

d) The speed of the astronaut

Explanation:

The sentence describes the speed of the astronaut. This speed value is 10meters per minute.

Now let us understand why;

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The value 10meters per minute clearly shows this instance. We do not know the direction the astronaut is moving towards.
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For example, 10m/s due west is a velocity value because we know the direction.

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3 years ago

How do Glaciers (located thousands of miles away) affect Florida's geography?

mr_godi [17]

Answer:

They affected ocean levels, causing waves to affect different areas of the coastline at different times. Lake Eola is a sinkhole lake.

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3 years ago

Define average velocity

maxonik [38]

Hello! mark me brainliest please

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3 years ago

Read 2 more answers

A satellite in outer space is moving at a constant velocity of 20.5 m/s in the +y direction when one of its on board thruster tu

Katyanochek1 [597]

Answer:

a) V_f = 25.514 m/s

b) Q =53.46 degrees CCW from + x-axis

Explanation:

Given:

- Initial speed V_i = 20.5 j m/s

- Acceleration a = 0.31 i m/s^2

- Time duration for acceleration t = 49.0 s

Find:

(a) What is the magnitude of the satellite's velocity when the thruster turns off?

(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.

Solution:

- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:

V_f = V_i + a*t

V_f = 20.5 j + 0.31 i *49

V_f = 20.5 j + 15.19 i

- The magnitude of the velocity vector is given by:

V_f = sqrt ( 20.5^2 + 15.19^2)

V_f = sqrt(650.9861)

V_f = 25.514 m/s

- The direction of the velocity vector can be computed by using x and y components of velocity found above:

tan(Q) = (V_y / V_x)

Q = arctan (20.5 / 15.19)

Q =53.46 degrees

- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.

4 0

3 years ago

A particle moves along a straight line. Its position at any instant is given by x = 32t− 38t^3/3 where x is in metre and t in se

Rudik [331]

Answer:

The acceleration of the object is -69.78 m/s²

Explanation:

Given;

postion of the particle:

$x = 32t - 38\frac{t^3}{3} \\\\$

The velocity of the particle is calculated as the change in the position of the particle with time;

$v = \frac{dx}{dt} = 32 - 38t^2\\\\when \ the \ particle \ is \ at \ rest, \ v = 0\\\\32-38t^2 = 0\\\\38t^2 = 32\\\\t^2 = \frac{32}{38} \\\\t = \sqrt{\frac{32}{38} } \\\\t = 0.918 \ s$

Acceleration is the change in velocity with time;

$a = \frac{dv}{dt} = -76t\\\\recall , \ t = 0.918 \ s\\\\a = -76(0.918)\\\\a = -69.78 \ m/s^2$

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2 years ago