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Mashutka [201]
3 years ago
15

Identify which objects will accelerate to the left, which will accelerate to the right,and those that will not accelerate. Optio

ns may be used multiple times or not at all.

Physics
1 answer:
Licemer1 [7]3 years ago
3 0

Answer:

The situation given in option A and B are examples for moving an object toward left side while the option C and option D are examples for moving an object toward right side. Option B will also be an example for not moving the object.

Explanation:

As per the option A statement, the force acting towards left is greater than the force acting toward right side. So the net force will be towards the direction having maximum magnitude. Thus, the box will move toward left side in option A. The same situation arises for the object in option B. But here the difference in the forces is only 1 N, so the change in the position of the object will be very less. Thus it may look like there is no acceleration in the box of option B.

Similarly, the force acting on the objects given in option C and D have magnitude greater towards the right side than towards the left side.  So these two will be accelerated toward the right side.

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a cannonball is fired with a speed of 76 m/s from the top of a cliff. It strikes the plane below with a speed of 89 m/s. if we n
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Answer:

we use the formula,

v {}^{2}  = u {}^{2}  + 2gh

89 {}^{2}  = 76 {}^{2}  + 2(10)h

h = (89 {}^{2}  - 76 {}^{2} ) \div 20

h= 107 m

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A 1.50-m string of weight 0.0125 N is tied to the ceil- ing at its upper end, and the lower end supports a weight W. Ignore the
Elena L [17]

The wave equation is missing and it is y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Answer:

A) 0.0534 seconds

B) 0.67N

C) 41

D) (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

Explanation:

we are given weight of string = 0.0125N

Thus, since weight = mg

Then, mass of string = 0.0125/9.8

Mass of string = 1.275 x 10⁻³ kg

Length of string; L= 1.5 m .

mass per unit length; μ = (1.275 x 10⁻³)/1.5

μ = 0.85 x 10⁻³ kg/m

We are given the wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Now if we compare it to the general equation of motion of standing wave on a string which is:

y(x,t) = Acos(Kx − ω t)

We can deduce that

angular velocity;ω = 4830 rad/s

Wave number;k = 172 rad/m

A) Velocity is given by the formula;

V = ω/k

Thus, V = 4830/172 m/s

V = 28.08 m /s

Thus time taken to go up the string = 1.5/28.08 = 0.0534 seconds

B) We know that in strings,

V² = F/μ

Where μ is mass per unit length and V is velocity.

Thus, F = V²*μ =28.08² x 0.85 x 10⁻³

F = 0.67N

C) Formula for wave length is given as; wave length;λ = 2π /k

λ = 2 x π/ 172

λ = 0.0365 m

Thus, number of wave lengths over whole length of string

= 1.5/0.0365 = 41

D) The equation for waves traveling down the string

= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

8 0
3 years ago
Two cars next to each other start moving in the same direction from rest, car A accelerates at m/s2 4 for 10 s and then travels
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If 3.61 m3 of a gas initially at STP is placed under a pressure of 2.67 atm , the temperature of the gas rises to 37.9 ∘C. What
Pavel [41]

Answer: 1.54m^3

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

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P_2 = final pressure of gas = 2.67 atm

V_1 = initial volume of gas = 3.61m^3

V_2 = final volume of gas = ?

T_1 = initial temperature of gas at STP = 0^oC=273+0=273K

T_2 = final temperature of gas = 37.9^oC=273+37.9=310.9K

Now put all the given values in the above equation, we get:

\frac{1atm\times 3.61m^3}{273K}=\frac{2.67\times V_2}{310.9K}

V_2=1.54m^3

Thus the final volume will be 1.54m^3

7 0
3 years ago
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