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3 years ago

Does anyone wanna be my friend????

Physics

2 answers:

Alla [95]3 years ago

7 0

// Okay, I would love to be your friend :D //

Marrrta [24]3 years ago

4 0

Answer:

Yes I will be your friend no problem bro

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A train travelling at 50km/h approaches another train moving towards the first at 90km/h. If they are 35km apart (on a straight

Ede4ka [16]

it will take 36 seconds

8 0

3 years ago

Which best describes how geothermal energy is used to make electricity?. . A.. Rocks are split to release energy that drives a s

In-s [12.5K]

Answer is b that is Heat energy from below the ground converts water to steam to drive a steam turbine attached to an electrical generator.. .

6 0

3 years ago

A steel ball of mass 0.500 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released whe

Liula [17]

Answer:

a) v₁fin = 3.7059 m/s (→)

b) v₂fin = 1.0588 m/s (→)

Explanation:

a) Given

m₁ = 0.5 Kg

L = 70 cm = 0.7 m

v₁in = 0 m/s ⇒ Kin = 0 J

v₁fin = ?

hin = L = 0.7 m

hfin = 0 m ⇒ Ufin = 0 J

The speed of the ball before the collision can be obtained as follows

Einitial = Efinal

⇒ Kin + Uin = Kfin + Ufin

⇒ 0 + m*g*hin = 0.5*m*v₁fin² + 0

⇒ v₁fin = √(2*g*hin) = √(2*(9.81 m/s²)*(0.70 m))

⇒ v₁fin = 3.7059 m/s (→)

b) Given

m₁ = 0.5 Kg

m₂ = 3.0 Kg

v₁ = 3.7059 m/s (→)

v₂ = 0 m/s

v₂fin = ?

The speed of the block just after the collision can be obtained using the equation

v₂fin = 2*m₁*v₁ / (m₁ + m₂)

⇒ v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)

⇒ v₂fin = 1.0588 m/s (→)

7 0

3 years ago

Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

djverab [1.8K]

Answer:

$v_{2}$ will be less than $v_{1}$ and $P_{2}$ will be greater than $P_{1}$ .

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass ( $m_{1}$ ) is entering in a system is equal to the rate at which the same amount of fluid mass ( $m_{2}$ ) is leaving the system.

Rate of mass flow can be written as,

$m = \rho A v$

where $\rho$ is the density of the fluid, $A$ is the area through which the fluid is flowing and $v$ is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas through which the fluid is passing and $v_{1}$ and $v_{2}$ are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, $A_{2} > A_{1}$ , so from the above formula $v_{2} < v_{1}$ .

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case $v_{2} < v_{1}$ the pressure in the large cross-sectional area $P_{2}$ will be greater than the pressure $P_{1}$ in the small cross sectional area, i.e.,

$P_{2} > P_{1}$ .

6 0

3 years ago

Help!! ASAP plz

VashaNatasha [74]

Can you include an image of the object and it’s dimensions?

4 0

3 years ago