what is the pressure from made from a book with the dimensions of 47 cm *86 cm and a measured force of 900N?

An object weighs 10N in air and 70N in water .What is it's weight when immersed in a liquid of relative density 1.5?

Morgarella [4.7K]

<h3>Answer:</h3>

5.5 N

<h3>Explanation:</h3>

<u>We are given;</u>

Real weight of an object in air as 10 N
Apparent weight in water as 7 N
Relative density of a liquid is 1.5

We need to calculate the apparent weight when in liquid .

First we calculate upthrust

Upthrust = Real weight - Apparent weight

= 10 N - 7 N

= 3 N

Then calculate the upthrust in the liquid.

we need to know that;

Relative density of a liquid = Upthrust in liquid/Upthrust in water

Therefore;

1.5 = U ÷ 3 N

Upthrust in liquid = 3 N × 1.5

= 4.5 N

Therefore, the upthrust of the object in the liquid is 4.5 N

But, Upthrust = Real weight - Apparent weight

Therefore;

Apparent weight in Liquid = Real weight - Upthrust

= 10 N - 4.5 N

= 5.5 N

Thus, the weight when the object is immersed in the liquid is 5.5 N

5 0

3 years ago

The BEST example of diffraction is the image of

Lady bird [3.3K]

The answer is C. Hope it helps

6 0

4 years ago

Read 2 more answers

A cannonball is catapulted toward a castle. The cannonball’s velocity when it leaves the catapult is 51.6 m/s at an angle of 37.

Andrei [34K]

Answer:

$v_{y}=35.21m/s$

Explanation:

From the exercise we know the cannonball's <u>initial velocity</u>, the <u>angle</u> which its released with respect to the horizontal and its <u>initial height</u>

$v_{o}=51.6m/s\\\beta =37.0º\\y_{o}=7m$

If we want to know whats the <u>y-component of velocity</u> we need to use the following formula:

$v_{y}^2=v_{oy}^2+ag(y-y_{o})$

Knowing that $g=-9.8m/s^2$

$v_{y}=\sqrt{((51.6m/s)sin(37))^2-2(9.8m/s^2)(0m-7m)}=35.21m/s$

So, the cannonball's y-component of velocity is $v_{y}=35.21m/s$

6 0

3 years ago

Convert 66.2 mi/h to m/s. 1 mi = 1609 m.<br> Answer in units of m/s.

Kryger [21]

Answer:

Explanation:

66.2mi/h to m/s

$\frac{1609m}{1mi}$ × $\frac{66.2mi}{1h}$ × $\frac{1h}{60min}$ × $\frac{1min}{60s}$

mi will go with mi
h with h
min with min
m/s is left

$\frac{1609*66.2}{60*60}$ = 29.58772222

Pls make sure of the way of solving to be sure...

4 0

4 years ago

(5) A 4 kg. object rests on a flat, horizontal surface with a static

baherus [9]

Answer:

F = 9.81 [N]

Explanation:

To solve this problem we must use Newton's third le which tells us that the sum of forces on a body that remains static must be equal to one resulting from these forces in the opposite direction.

Let's perform a summation of forces on the vertical axis-y to determine the normal force N.

∑F = 0 (axis-y)

$N - m*g = 0$

where:

m = mass = 4 [kg]

g = gravity acceleration = 9.81 [m/s²]

$N - (4*9.81)=0\\N = 39.24 [N]$

Now we know that the frictional force can be calculated using the following equation.

f = μ*N

where:

f = friction force [N]

μ = friction coefficient = 0.25

N = normal force = 39.24 [N]

Now replacing:

$f = 0.25*39.24\\f = 9.81[N]$

Then we perform a sum of forces on the X-axis equal to zero. This sum of forces allows us to determine the minimum force to be able to move the object in a horizontal direction.

∑F = 0 (axis-x)

$F-f=0\\F-9.81=0\\F= 9.81[N]$

If the coefficient was smaller, a smaller force (F) would be needed to start the movement, this can be easily seen by replacing the value of 0.25, by smaller values, such as 0.1 or 0.05.

If the coefficient were larger, a larger force would be needed.

3 0

3 years ago