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2 years ago

Please help me!

Physics

1 answer:

sergeinik [125]2 years ago

6 0

C. The Densities are equal.

<h3>What is density?</h3>

Density is mass per unit volume or mass of a unit volume of a material substance.

If m1, V1 and D1 = mass, volume and density respectively of ball C

m2, V2 and D2 = mass, volume and density respectively of ball D

According to the Question ,

$V_{1} = 3V_{2} , m_{2} = \frac{1}{3} (m_{1} ) \\ \\= m_{1} = 3m_{2}$

Therefore,

$\frac{D_{1} }{D_{2} } = (\frac{m_{1} }{V_{1} } )* (\frac{m_{2} }{V_{2} } )\\ \\= (\frac{3m_{2} }{3V_{2} })*(\frac{V_{2} }{m_{2} }) \\\\= 1$

Hence, D1 = D2

Learn more about density here:brainly.com/question/15164682

#SPJ1

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Where and when does the sun stay up for 24hours

DIA [1.3K]

Answer:

abisko, sweden

Explanation:

A bisco is the home to their eyes sky station an epic center for Aurora expanses and northern Sweden. During summer months, the Sun Bates to town and up to 24 hours of sunlight per day.

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3 years ago

Would a ship in Lake Ontario (fresh water) float higher or lower in the water than in the Atlantic Ocean (salt water)?

ohaa [14]

Atlantic Ocean (salt water)

8 0

2 years ago

Consider an electron, of charge magnitude e = 1.602 10-19 C and mass me = 9.11 10-31 kg, moving in an electric field with an ele

Tems11 [23]

Answer: v= 7.509 x 10^6 m/s

B) the lower plate because the electron is negatively charged

Explanation:

From the question

Electronic charge (q) =1.602 x 10^-19c

Electric field intensity (E) = 8 x 10² = 800N/C

Mass of electron (m) = 9.11 x 10^-31 kg

Length of plate (L) = 50cm=0.5m

Distance between plates (D) = 20cm = 0.2m

Since the electron is entering a uniform electric field, the resulting motion will be of a constant acceleration and can be defined by the equations of motion with a constant acceleration.

From newton's law of motion.

F= ma

The force (F) is coming from the electric field which is

F=Eq.

Thus F = 800 x 1.602 x 10^-19

F = 1.2816 x 10^-16 N

Acceleration of electron (a) = F/m where m is the mass of electron (given above)

Hence

a = 1.2816 x 10^-16 / 9.11 x 10^-31

a= 1.41 x 10¹⁴ m/s².

Using Newton laws of motion to get velocity, we recall that v²= u² + 2ad

Where v is final velocity, u is initial velocity (zero in this case because the electron starts it motion from rest), a= acceleration, d= distance traveled (which in this case is distance between plates)

v² = 0² + 2(1.41 x 10¹⁴) x 0.2

v² = 5.64 x 10¹³

Thus v = √ 5.64 x 10¹³

v = 7.509 x 10^6 m/s

Since the electric field is upwards it denotes that the positive plate is downward and the negative is upward ( this is because electric flux from a positive charge has an outward flow and that from a negative charge has an inward flow. So from our questions, if the electric field is upward, it means it is starting from the bottom plate which will be positive) the electron (which is negatively charged) will be attracted to the positive plate which is downward for this question of ours.

So therefore, the electron is attracted to the downward plate because it (electron) is negative

Option B

4 0

3 years ago

Select the correct answer.

Triss [41]

Answer:

Jim's kinetic energy is 54.67 J.

Explanation:

Given that,

Mass, m = 15 kg

Velocity, V = 2.7 m/s

We need to find the Jim's kinetic energy. We know that when the object is in motion, it has kinetic energy. This energy is given by :

$E=\dfrac{1}{2}mv^2$

$E=\dfrac{1}{2}\times 15\ kg\times (2.7\ m/s)^2$

E = 54.67 J

So, Jim's kinetic energy is 54.67 J. Hence, this is the required solution.

5 0

3 years ago

The force P is applied to the 45-kg block when it is at rest. Determine the magnitude and direction of the friction force exerte

cluponka [151]

Answer:

Check attachment for free body diagram of the question.

I used the free body diagram and the angles given in the diagram missing in the question above, but I used the data given in the above question.

Explanation:

Let frictional force be Fr acting down the plane

Let analyze the structure before inserting values

Using Newton's second law along the y-axis

ΣFy = Fnet = m•ay

Since the body is not moving in the y-direction, then ay = 0

N+PSinβ — WCosθ = 0

N+PSin20—441.45Cos15 = 0

N+PSin20—426.41 = 0

N = 426.41 — PSin20 , equation 1

The maximum Frictional force to be overcome is given as

Fr(max) = μsN

Fr(max) = 0.25(426.41 — PSin20)

Fr(max)= 106.6 —0.25•PSin20

Fr(max) = 106.6 — 0.08551P, equation 2

This is the maximum force that must be overcome before the body starts to move

Using Newton's law of motion in the x direction

Note, we took the upward direction up the plane as the direction of motion since the force want to move the block upward

Fnetx = ΣFx

Fnetx = P•Cosβ —W•Sinθ — Fr

Fnetx = P•Cos20—441.45•Sin15—Fr

Fnetx = 0.9397P — 114.256 — Fr

Equation 3

When Fnetx is positive, then, the body is moving up the plane, if Fnetx is negative, then, the maximum frictional force has not yet being overcome and the object is still i.e. not moving

a. When P = 0

From equation 2

Fr(max) = 106.6 — 0.08551P

Fr(max) = 106.6 — 0.08551(0)

Fr(max)= 106.6 N

So, 106.6N is the maximum force to be overcome

So, here the only force acting on the body is the weight and it acting down the plane, trying to move the body downward.

Wx = WSinθ

Wx = 441.45× Sin15

Wx = 114.256 N.

Since the force trying to move the body downward is greater than the maximum static frictional force, then the body is not in equilibrium, it is moving downward.

So, finding the magnitude of frictional force

From equation 1

N = 426.41 — PSin20 , equation 1

N = 426.41 N, since P=0

Then, using law of kinetic friction

Fr = μk • N

Fr = 0.22 × 426.41

Fr = 93.81 N.

b. Now, when P = 190N

From equation 2

Fr(max) = 106.6 — 0.08551(190)

Fr(max) = 106.6 —16.2469

Fr(max)= 90.353 N

So, 90.353 N is the maximum force to be overcome

Now the force acting on the x axis is the horizontal component of P and the horizontal component of the weight

Fnetx = P•Cosβ —W•Sinθ

Fnetx = 190Cos20 — 441.45Sin15

Fnetx = 64.29N

So the force moving the body up the incline plane is 64.29N

Fnetx < Fr(max)

Then, the frictional force has not being overcome yet.

Then, the body is in equilibrium.

Then, applying equation 3.

Fnetx = 0.9397P — 114.256 — Fr

Fnetx = 0, since the body is not moving

0 = 0.9397(190) —114.246 — Fr

Fr = 64.297 N

Fr ≈ 64.3N

c. When, P = 268N

From equation 2

Fr(max) = 106.6 — 0.08551(268)

Fr(max) = 106.6 —16.2469

Fr(max)= 83.68 N

So, 83.68 N is the maximum force to be overcome

Now the force acting on the x axis is the horizontal component of P and the horizontal component of the weight

Fnetx = P•Cosβ —W•Sinθ

Fnetx = 268Cos20 — 441.45Sin15

Fnetx = 137.58 N

So the force moving the body up the incline plane is 137.58 N

Fnetx > Fr(max)

Then, the frictional force has being overcome.

Then, the body is not equilibrium.

So, finding the magnitude of frictional force

From equation 1

N = 426.41 — 268Sin20 , equation 1

N = 334.75 N, since P=268N

Then, using law of kinetic friction

Fr = μk • N

Fr = 0.22 × 334.75

Fr = 73.64 N

d. The required force to initiate motion is the force when the block want to overcome maximum frictional force.

So, Fnetx = Fr(max)

Px — Wx = Fr(max)

From equation 1

Fr(max) = 106.6 — 0.08551P,

P•Cosβ-W•Sinθ = 106.6 — 0.08551P

P•Cos20 — 441.45•Sin15 = 106.6 — 0.08551P

P•Cos20—114.256=106.6 - 0.08551P

PCos20+0.08551P =106.6 + 114.256

1.025P=220.856

P = 220.856/1.025

P = 215.43 N

3 0

3 years ago