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2 years ago

A body displaces 50cm^3 of water. What is the up-thrust on the body?

Physics

1 answer:

SVEN [57.7K]2 years ago

3 0

Answer:

Force of buoyancy is given by

Fb = Weight of displaced water

= mg

But, m=density x volume

= ρV g

where ρ = density of the water

V = volume of displaced water = volume of block

g = acceleration due to gravity

Given that,

V = 50 cm3

g = 981 cm/s2

ρ = 1 g/cm3

Fb = 1*50*981

= 49050 dyne

= 0.49 N

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A plane travels down a runway 2750 m before it lifts off at an angle of 37 degrees from the horizontal. The plane has traveled 1

ss7ja [257]

Answer: 4.236km

Explanation:

Let's define the point (x, y) as:

x = horizontal distance moved.

y = vertical distance moved.

If the plane starts in the point (0, 0) then:

"A plane travels down a runway 2750 m before it lifts off..."

At this time, the position will be:

P = (0 + 2750m, 0) = (2750m, 0).

"it lifts off at an angle of 37 degrees from the horizontal. The plane has traveled 1.8 km since its wheels left the ground."

In this case, as the angle is measured from the horizontal, the components will be:

x = 1.8km*cos(37°) = 1.438km

y = 1.8km*sin(37°) = 1.083 km

Then the new position is:

P = (2750m + 1.438 km, 0 + 1.083 km)

Let's write it using the same units for all the quantities:

we know that

1km = 1000m

Then:

2750m = (2750/1000) km = 2.750 km.

Then we can write the new position as:

P = (2.750 km + 1.438km, 1.083km) = (4.188km, 1.083km)

Now, we define the displacement as the distance between the final position and the initial position.

The distance between two points (a, b) and (c, d) is:

D = √( (a c)^2 + (b - d)^2)

In this case the points are:

(0, 0) for the initial position

(4.188km, 1.083km) for the final position.

And the displacement will be:

D = √( (4.188km - 0)^2 + (1.083 - 0)^2) = 4.236km

5 0

3 years ago

A body with the inertial

Andrews [41]

Answer:

Explanation:

Hi there,

To get started, recall the kinematic equations from either a textbook, equation sheet, etc. Kinematic equations are used when acceleration is <em>constant,</em> as stated in the prompt.

Best way to use kinematic equations is to see which variable you are looking for, then which variable is unknown to you and is not needed for that equation.

a) average velocity

Takes the form of:

$v_a_v_g=\frac{d_t_o_t_a_l}{t}=\frac{v+v_0}{2}$ this is the literal definition of average velocity; initial plus final divided by 2.

We know total displacement and total time elapsed, so we will use the middle form of the equation:

$v_a_v_g=\frac{1640m}{40s}=41 \ m/s$

b) the final velocity

We can still use the average velocity formula, as the other two equations that include final velocity have acceleration variable which is unknown as of now.

Solve for final velocity:

$v=(2v_a_v_g)-v_o = 2(41 \ m/s) - (8 m/s) = 74 m/s\\$ this makes sense, since a velocity later in time is higher than a velocity earlier in time. It is increasing with increasing time because of acceleration.

c) the acceleration

There are two equations that can be used to solve this, but we will use the less time-consuming one, but both produce same answer:

$a = \frac{v-v_0}{t_t_o_t_a_l} = \frac{(74-8)m/s}{40s} =1.65 m/s^{2}$

Notice, change in velocity over change in time, and acceleration is constant. When acceleration is constant, it models a linear function, and acc. is just slope!

Study well and persevere. If you liked this solution, hit Thanks or give a rating!

thanks,

3 0

3 years ago

Why is radiation fog more likely near the center of an anticyclone than near the center of a cyclone?

Zigmanuir [339]

Radiation fog is the fog that is formed when the heat absorbed the Earth's surface is released into the atmosphere producing fog. This only occurs when the air is clear and calm. In the center of an anticyclone, the conditions of the air are clear and calm which is favorable for the formation of radiation fog. The center of cyclones, on the other hand, is turbulent and cloudy which prevents the formation of radiation fogs.

4 0

3 years ago

This lab is investigating the relationship between mass, ________, and momentum.

Sav [38]

This lab is investigating the relationship between mass, <u>Speed </u>, and momentum.

Momentum is manufactured from the mass and speed of an object. it's miles a vector quantity, owning a significance and a direction. If m is an object's mass and v is its speed, then the object's momentum is p.

Momentum in an easy way is a quantity of movement. right here amount is measurable because if an item is moving and has mass, then it has momentum. If an object no longer flows then it has no momentum. however, in regular existence, it has an important but many people didn't understand it.

Momentum gives the connection between the mass, pace, and direction of an object. Any exchange in momentum results in pressure. So, an exchange in momentum is used to determine the force appearing upon the item.

Learn more about momentum here:-brainly.com/question/1042017

#SPJ1

7 0

1 year ago

True or False: Objects that have more mass also have more gravity.

Fantom [35]

Answer:

True

Explanation:

The smaller object usually moves towards the larger object, because the larger object has more mass than the smaller object, so it has a larger gravitational pull.

7 0

3 years ago

Read 2 more answers

A body displaces 50cm^3 of water. What is the up-thrust on the body? ​

A body displaces 50cm^3 of water. What is the up-thrust on the body?