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mojhsa [17]
2 years ago
6

An empty glass beaker has a mass of 103 g. When filled with water, it has a total mass of 361g.

Physics
1 answer:
aivan3 [116]2 years ago
7 0

Answer:

0.96 gcm¯³

Explanation:

From the question given above, the following data were obtained:

Mass of empty beaker = 103 g

Mass of beaker + water = 361 g

Mass of beaker + oil = 351 g

Density of water = 1 gcm¯³

Density of cooking oil =?

Next, we shall determine the mass of water. This can be obtained as follow:

Mass of empty beaker = 103 g

Mass of beaker + water = 361 g

Mass of water =?

Mass of water = (Mass of beaker + water) – (Mass of empty beaker)

Mass of water = 361 – 103

Mass of water = 258 g

Next, we shall determine the volume of the beaker. This can be obtained by calculating the volume of water in the beaker.

Density of water = 1 gcm¯³

Mass of water = 258 g

Volume of water =?

Density = mass /volume

1 = 258 / volume

Cross multiply

1 × volume = 258

Volume of water = 258 cm³

Thus the volume of the beaker is 258 cm³.

Next, we shall determine the mass of the cooking oil. This can be obtained as follow:

Mass of empty beaker = 103 g

Mass of beaker + oil = 351 g

Mass of cooking oil =?

Mass of cooking oil = (Mass of beaker + oil) – (Mass of empty beaker)

Mass of cooking oil = 351 – 103

Mass of cooking oil = 248 g

Finally, we shall determine the density of the cooking oil. This can be obtained as follow:

Mass of cooking oil = 248 g

Volume of the beaker = 258 cm³

Density of cooking oil =?

Density = mass / volume

Density = 248 / 258

Density of cooking oil = 0.96 gcm¯³

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Answer:

B. 4 m/s

Explanation:

v=d/t

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2 years ago
A ball is thrown with a velocity of 35 meters per second at an angle of 30° above the horizontal. which quantity has a magnitude
enot [183]
The quantity that has a magnitude of zero when the ball is at the highest point in its trajectory is the vertical velocity.

In fact, the motion of the ball consists of two separate motions:
- the horizontal motion, on the x-axis, which is a uniform motion with constant velocity v_x=v_0 cos 30^{\circ}, where v_0=35 m/s
- the vertical motion, on the y-axis, which is a uniformly accelerated motion with constant acceleration g=9.81 m/s^2 directed downwards, and with initial velocity v_y=v_= sin 30^{\circ}. Due to the presence of the acceleration g on the vertical direction (pointing in the opposite direction of the initial vertical velocity), the vertical velocity of the ball decreases as it goes higher, up to a point where it becomes zero and it reverses its direction: when the vertical velocity becomes zero, the ball has reached its maximum height. 
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2 years ago
If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g
yulyashka [42]

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178200 g mile pounds

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Work= 165 pounds *g* 1080 m=  178200 g mile pounds

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2 years ago
A 60-μC charge is held fixed at the origin and a −20-μC charge is held fixed on the x axis at a point x = 1.0 m. If a 10-μC char
Aleksandr [31]

Answer:

Ek =  8,79 [J]

Explanation:

We are going to solve this problem, using  the energy conservation principle

State 1 or initial state (charges at rest t=0)

E₁  = Ek  + U₁

As charge are at rest Ek = 0

And  U₁ has two components

U₁₂   = K * Q₁*Q₂ / 0,4          and    U₃₂  = K*Q₃*Q₂ / 0,6

U₁₂  = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,4  ⇒ U₁₂ = 9*60*10*10⁻³/0,4

U₃₂ =  - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,6  ⇒ U₃₂ = - 9*20*10*10⁻³/0,6

U₁₂ = 540*10⁻2/0,4 [J]   ⇒13,5 [J]

U₃₂ = - 180*10⁻² /0,6 [J] ⇒ - 3 [J]

Then   E₁ = E₁₂ + E₃₂    

E₁ = 10,5 [J]

At  the moment of Q₂ passing x = 40 cm  or 0,4 m

E₂ = Ek + U₂

We can calculate the components of U₂ in this new configuration

U₂  =  U₁₂  + U₃₂

U₁₂  = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,7   ⇒  U₁₂ = 9*60*10*10⁻³/0,7

U₁₂ = 540*10⁻²/0,7       U₁₂ = 7,71 [J]

U₃₂ =  - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,3  ⇒ U₃₂ = -  9*20*10*10⁻³/0,3

U₃₂ = -  9*20*10⁻²/0,3  

U₃₂ = - 6

U₂ = 7,71 -6

U₂ = 1,71 [J]

Then as  

E₂  = Ek + U₂  and  E₂ = E₁

Then

Ek + U₂ = E₁

Ek =  10,5 - U₂    

Ek  = 10,5 - 1,71

Ek =  8,79 [J]

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1 year ago
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1) weight of the box: 980 N

The weight of the box is given by:

W=mg

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W=(100.0 kg)(9.8 m/s^2)=980 N


2) Normal force: 630 N

The magnitude of the normal force is equal to the component of the weight which is perpendicular to the ramp, which is given by

N=W cos \theta

where W is the weight of the box, calculated in the previous step, and \theta=50^{\circ} is the angle of the ramp. Substituting, we find

N=(980 N)(cos 50^{\circ})=630 N


3) Acceleration: 7.5 m/s^2

The acceleration of the box along the ramp is equal to the component of the acceleration of gravity parallel to the ramp, which is given by

a_p = g sin \theta

Substituting, we find

W_p = (9.8 m/s^2)(sin 50^{\circ})=7.5 m/s^2

5 0
2 years ago
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