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2 years ago

Calculate the density of mercury if 500 cm3

Physics

1 answer:

notka56 [123]2 years ago

5 0

Answer:

The density of the mercury is 13.2 g/cm³

Explanation:

Density is a measurement that compares the amount of matter an object

has to its volume

Density is equal to mass divided by volume

We need to find the density of mercury if 500 cm³ has a mass of

6.60 kg in g/cm

We must to change The kilogram to grams

The mass of mercury is 6.60 kilograms

1 kilogram = 1000 grams

6.60 kilograms = 6.60 × 1000 = 6600 grams

Density = mass ÷ volume

The volume of the mercury is 500 cm³

The density = 6600 ÷ 500

The density = 13.2 g/cm³

The density of the mercury is 13.2 g/cm³

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A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then

Elan Coil [88]

Answer:

A) $W_{ff} =-744.12J$

B) $F_f=-W_{ff}*sin\theta /hy = 112.75N$

C) $F_{f2}=207.58N$

Explanation:

This question is incomplete. The full question was:

A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s.

Part (a) Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement.

Part (b) The ramp makes an angle θ with the ground, where θ = 30°. Write an expression for the magnitude of the friction force, fr, between the ramp and the skateboarder.

Part (c) When the skateboarder reaches the bottom of the ramp, he continues moving with the speed vf onto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force, Fgrass in newtons, between the skateboarder and the grass.

For part A), we make a balance of energy to calculate the work done by the friction force:

$W_{ff}=\Delta E$

$W_{ff}=1/2*m*vf^2-m*g*hy$

$W_{ff}=-744.12J$

For part B), we use our previous value for the work:

$W_{ff}=-F_f*(hy/sin\theta)$ Solving for friction force:

$F_f=-W_{ff}*sin\theta /hy$

$F_f=112.75N$

For part C), we first calculate the acceleration by kinematics and then calculate the module of friction force by dynamics:

$Vf^2=Vo^2+2*a*d$

Solving for a:

$a=-3.844m/s^2$

Now, by dynamics:

$|F_f|=|m*a|$

$|F_f|=207.58N$

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2 years ago

You throw a baseball at an angle of 30.0∘∘ above the horizontal. It reaches the highest point of its trajectory 1.05 ss later. A

garri49 [273]

Answer:

The speed with which the baseball leaves the hand = 20.58 m/s

Explanation:

The time take to reach highest height during a projectile's flight is given by

t = (u sin θ)/g

u = initial velocity of the baseball = ?

θ = angle of throw above the horizontal

g = acceleration due to gravity = 9.8 m/s²

1.05 = (u sin 30)/9.8

u = (1.05 × 9.8)/0.5

u = 20.58 m/s

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2 years ago

A block of mass 2.00 kg is initially at rest at x=0 on a slippery horizontal surface for which there is no friction. Starting at

Allisa [31]

Answer:

x = 1,185 m , t = 4/3 s , F = - 4 N

Explanation:

For this exercise we use Newton's second law

F = m a = m dv /dt

β - α t = m dv / dt

dv = (β – α t) dt

We integrate

v = β t - ½ α t²

We evaluate between the lower limits v = v₀ for t = 0 and the upper limit v = v for t = t

v-v₀ = β t - ½ α t²

the farthest point of the body is when v = v₀ = 0

0 = β t - ½ α t²

t = 2 β / α

t = 2 4/6

t = 4/3 s

Let's find the distance at this time

v = dx / dt

dx / dt = v₀ + β t - ½ α t2

dx = (v₀ + β t - ½ α t2) dt

We integrate

x = v₀ t + ½ β t - ½ 1/3 α t³

x = v₀ 4/3 + ½ 4 (4/3)² - 1/6 6 (4/3)³

The body comes out of rest

x = 3.5556 - 2.37

x = 1,185 m

The value of force is

F = β - α t

F = 4 - 6 4/3

F = - 4 N

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1 year ago

The rear wheel on a clown’s bicycle has twice the radius of the front wheel. (a) When the bicycle is moving, is the linear speed

8_murik_8 [283]

Answer:

a). same as

b). less than

Explanation:

a). When a bicycle is moving, the linear speed at the top of the rear wheel is same as the linear speed at the top of the front wheel. Since the clown's bicycle is a rigid body, both the wheels that is the front wheel and the rear wheel will move with the same linear speed.

b). Since we know that angular speed varies inversely to the radius of the wheel.

That is ω = 1 / r

Since the rear wheel has twice the radius of that of the front wheel, therefore real wheel will have less angular speed than the front wheel.

Therefore, the angular speed of the rear wheel is less than the angular speed of the front wheel.

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1 year ago

A transverse wave is traveling on a string stretched along the horizontal x-axis. The equation for the

maxonik [38]

Answer:

A) 0.33 m/s

Explanation:

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y = a cos ( ω t − kx ) , k = 2 π / λ

Amplitude, a = 0.002 m

Wavenumber (k)=47.12 and wavelength ( λ ) = 0.133 m

Time period(T)=0.0385 s and angular frequency ( ω ) = 52 π rad/s

Maximum speed of the string is given by aw

Therefore ; max. speed = 0.002 x 52 π = 0.327 m/s

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2 years ago