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3 years ago

PLEASE HELP ME ASAPPPPPP

Physics

2 answers:

olganol [36]3 years ago

6 0

Answer:

i cants see it

Explanation:

Helen [10]3 years ago

5 0

Answer:

a drummer practicing a drum set

Explanation:

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Two bicycle tires are set rolling with the same initial speed of 4.00 m/s along a long, straight road, and the distance each tra

77julia77 [94]

Answer:

The coefficient of rolling friction for the tire under low pressure is 0.0342.

Explanation:

Two bicycle tires are set rolling with the same initial speed of 4.00 m/s

Final speed of both the bicycle, speed is reduced by half is measured, v = 2 m/s.

Here,

$u_kmg = ma\\\\a=\mu g$

Using third equation of motion as :

$v^2-u^2=2as\\\\v^2-u^2=2\mu gs\\\\\mu =\dfrac{v^2-u^2}{2gs}\\\\\mu =\dfrac{4^2-2^2}{2\times 9.8\times 17.9}\\\\\mu=0.0342$

So, the coefficient of rolling friction for the tire under low pressure is 0.0342.

5 0

3 years ago

An elevator is moving downward when someone presses the emergency stop button. The elevator comes to rest a short time later. Gi

OLga [1]

Answer:

<em>Answer: - , + </em>

Explanation:

<u>Velocity and Acceleration </u>

In general terms, velocity and acceleration in the plane are both vectors with two components. They don't have 'sign' but magnitude and direction. If we know motion is restricted to a simple line, then we can define the positive and negative references.

In vertical motion, the positive reference is commonly pointed upwards, but if we chose the opposite way, the results shouldn't change in magnitude, just their signs. Let's choose like proposed, the positive direction upwards.

Since the elevator was moving downwards, its velocity is negative. When the button is pushed, it continues to move downwards, but slower. Its velocity is still negative.

The acceleration has the same sign as the velocity if it tends to increase the magnitude of the object, and is negative in the opposite case. Let's look at the formula

$\displaystyle a=\frac{v_f-v_o}{t}$

In this case, $v_f=0$ because the elevator will eventually stop, and $v_o$ is negative. It means the acceleration is

$\displaystyle a=\frac{-v_o}{t}$

It results in a positive value

$\boxed{Answer: -,+}$

Note: If we chose the opposite references, then the answer would be: +,-

5 0

3 years ago

Three crates with various contents are pulled by a force F pull = 3615 N across a horizontal, frictionless roller‑conveyor syste

Dmitry_Shevchenko [17]

Answer:

m1=914.9kg

m2=604.9kg

m3=864.75kg

Explanation

I think we are suppose to find the mass of the crate.

The effective force that moves the body in positive x direction is 3615N

ΣFx = Σma

Then Fx=3615N

Then the masses be m1, m2 and m3

Then,

ΣF = Σ(ma)

3615=(m1+m2+m3)a

Given that a=1.516

The masses are

m1+m2+m3=, 2384.56. Equation 1

Between mass 1 and mass 2 is, F12=1387.

The effective force that pull mass 1 is 1387.

F12=m1 ×a

Therefore,

m1=F12/a

m1=1387/1.516

m1=914.9kg.

The effective force that pulls crate 1 and crate 2 is F23

F23=(m1+m2)a

Therefore

2304=(m1+m2)a

Therefore, since a=1.516

m1+m2=2304/1.516

m1+m2=1519.8kg

Since m1=914.9kg

So, m2=1519.8-m1

m2=1519.8-914.9

m2=604.9kg

Also from equation 1

m1+m2+m3=2384.56

Since m1=914.9kg and m2=604.9kg

Then, m3=2384.56-604.9-914.9

m3=864.75kg

3 0

3 years ago

A 5 m3 tank containing 5kg of an unknown ideal gas at 500 kPa is connected through a valve to another tank containing 10 kg of t

Ivan

Answer:

a) $V_{T} = 9\,m^{2}$ , b) $m_{T} = 15\,kg$ , c) $P_{T} = 416.667\,kPa$

Explanation:

a) The equation of state for ideal gas is:

$P \cdot V = \frac{m}{M}\cdot R_{u}\cdot T$

Given the existence of an isothermal process, the following relation is derived:

$\frac{P_{1}\cdot V_{1}}{m_{1}} = \frac{P_{2}\cdot V_{2}}{m_{2}}$

The volume of the other tank is:

$V_{2} = \left(\frac{m_{2}}{m_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}}\right)\cdot V_{1}$

$V_{2} = \left(\frac{10\,kg}{5\,kg} \right)\cdot \left(\frac{200\,kPa}{500\,kPa}\right)\cdot (5\,m^{3})$

$V_{2} = 4\,m^{3}$

The total volume is:

$V_{T} = V_{1} + V_{2}$

$V_{T} = 5\,m^{3} + 4\,m^{3}$

$V_{T} = 9\,m^{2}$

b) The total mass is:

$m_{T} = m_{1} + m_{2}$

$m_{T} = 5\,kg + 10\,kg$

$m_{T} = 15\,kg$

c) The pressure of the gas in the two tanks is:

$P_{2} = \left(\frac{m_{2}}{m_{1}} \right)\cdot \left(\frac{V_{1}}{V_{2}}\right)\cdot P_{1}$

$P_{T} = \left(\frac{15\,kg}{5\,kg}\right)\cdot \left(\frac{5\,m^{2}}{9\,m^{2}} \right)\cdot (500\,kPa)$

$P_{T} = 416.667\,kPa$

3 0

4 years ago

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rjkz [21]

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4 years ago

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