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marissa
2 months ago
12

Students work together during an experiment about Newton’s laws. The students use a setup that consists of a cart of known mass

connected to one end of a string that is looped over a pulley of negligible friction, with its other end connected to a hanging mass. The cart is initially at rest on a horizontal surface. Students have access to common laboratory equipment to make measurements of components of the system. The students double the mass that hangs from the string. They also replace the original cart with a new cart that has double the mass. By doubling both masses, how will the tension in the string and the acceleration of the cart change?

Physics
1 answer:
Yuliya22 [3.3K]2 months ago
8 0

Answer:

The tension in the string will increase by a factor of two, while the acceleration will not change.

Explanation:

To approach this problem, we begin by sketching the situation and creating free body diagrams for both masses involved. (Refer to the attached image)

Let’s first examine the free body diagram for the cart. Assuming there’s no friction between the cart and the horizontal surface, and focusing solely on horizontal forces, we can write the force balance as follows:

\sum F_{1}=m_{1}a

Given that tension is the only horizontal force on the cart, we state:

T=m_{1}a

Next, consider the hanging mass, taking downward as the positive direction. Summing forces yields:

\sum F_{2}=m_{2}a

The two forces acting here are gravity and tension, so:

-T+m_{2}g=m_{2}a

Combining the two equations results in:

-m_{1}a+m_{2}g=m_{2}a

Solving for acceleration gives:

m_{1}a+m_{2}a=m_{2}g

a(m_{1}+m_{2})=m_{2}g

a=\frac{m_{2}g}{m_{1}+m_{2}}

This represents the acceleration for the original setup. What happens if both masses are doubled? Let’s analyze:

a_{double masses}=\frac{2m_{2}g}{2m_{1}+2m_{2}}

Factoring 2 out of the denominator produces:

a_{double masses}=\frac{2m_{2}g}{2(m_{1}+m_{2})}

Which simplifies to:

a_{double masses}=\frac{m_{2}g}{m_{1}+m_{2}}

You can observe this matches the original acceleration, confirming it remains unchanged.

Regarding tension, by substituting the doubled masses into the initial force expression:

T_{double}=2m_{1}a

The initial tension was:

T=m_{1}a

Therefore, doubling the masses causes the tension to double as well.

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Suppose Sammy Sosa hits a home run which travels 361. ft (110. m). Leaving the bat at 50 degrees above the horizontal, how high
Yuliya22 [3333]

Response:

The horizontal span of Sosa is 276.526 ft or 84.28 meters.

Explanation:

As illustrated in the diagram, let point O denote Sosa's starting position. She travels 361 ft at a 50-degree angle relative to the horizontal.

sin 50 = \frac{OM}{OP}

0.7660 = h / 361

h = 276.526 ft

h = 84.28 meters

The horizontal distance of Sosa is 276.526 ft or 84.28 meters.

5 0
1 month ago
An advertiser who calls a basement apartment a “garden apartment” is using the rhetorical device called
serg [3582]
I would choose B or D, based on the location. If it’s situated in South Kensington, London, then D might be appropriate. Conversely, if it’s in an underprivileged area, I'd opt for B.
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26 days ago
While dangling a hairdryer by its cord, you observe that the cord is vertical when the hairdryer isoff and, once it is turned on
Yuliya22 [3333]

Answer:

The air exiting from the hairdryer is moving at a speed of 10 m/s.

Explanation:

The thrust generated by the hairdryer enables it to maintain an elevation angled at 5° from vertical; thus, we derive from the force diagram

(1).\: tan (5^o) = \dfrac{F_t}{Mg}

by substituting M =0.420kg, g = 9.8m/s^2 into the equation and resolving for F_t we find:

F_t = Mg\:tan(5^o)

F_t = (0.420kg)(9.8m/s^2)\:tan(5^o)

\boxed{F_t = 0.3601N.}

This thrust is linked to the speed of air ejection v through the equation

(2).\: F_t = v\dfrac{dM}{dt}

where dM/dt signifies the rate of air ejection, which is known to be

0.06m^3/2s  = 0.03m^3/s

and since 1m^3 = 1.2kg,

0.03m^3/s \rightarrow 0.036kg/s

\dfrac{dM}{dt}  = 0.036kg/s,

by inserting these values into equation (2), we obtain the value of F_t as:

0.3601N = 0.036v

resulting in

v= \dfrac{0.3601}{0.036}

\boxed{v =10m/s.}

which indicates the air velocity discharged from the hairdryer.

6 0
2 months ago
A basketball center holds a basketball straight out, 2.0 m above the floor, and releases it. It bounces off the floor and rises
Softa [3030]

Answer:

a) The ball's velocity just prior to hitting the ground measures -6.3 m/s

b) The ball's velocity right after bouncing off the ground registers at 3.1 m/s

c) The average acceleration's magnitude is 470 m/s², and its direction is upward, forming a 90º angle with the ground.

Explanation:

To begin, let’s assess the time it takes for the ball to reach the floor:

The equation outlining the ball's position is:

y = y0 + v0 * t + 1/2 g * t²

Where:

y = position at given time t

y0 = initial position

v0 = initial velocity

t = time

g = acceleration triggered by gravity

We establish the ground as the reference origin.

a) Since the ball is released rather than thrown, the initial velocity v0 is 0. The direction of acceleration is downward, directed towards the origin; thus, “g” is treated as negative. When the ball contacts the ground, its position will be 0. Therefore:

0 = 2.0 m + 0 m/s *t - 1/2 * 9.8 m/s² * t²

-2.0 m = -4.9 m/s² * t²

t² = -2.0 m / - 4.9 m/s²

t = 0.64 s

The motion equation for a falling body is:

v = v0 + g * t      where "v" denotes the velocity

Since v0= 0:

v = g * t = -9.8 m/s² * 0.64 s = -6.3 m/s

b) The pebble's speed reaches 0 during its maximum height. To find the time taken for the pebble to achieve that height, we can use the velocity equation and then substitute that time in the position equation to derive the initial velocity:

v = v0 + g * t

0 = v0 + g * t

-v0/g = t

Replacing t in the position equation, knowing the maximum height is 1.5 m:

y = y0 + v0 * t + 1/2* g * t² y = 1.5 m y0 = 0 m t = -v0/g

1.5 m = v0 * (-v0/g) + 1/2 * g (-v0/g)²

1.5 m = - v0²/g - 1/2 * v0²/g

1.5 m = -3/2 v0²/g

1.5 m * (-2/3) * g = v0²

1.5 m * (-2/3) * (-9.8 m/s²) = v0²

v0 = 3.1 m/s

c) The average acceleration can be determined by:

a = final velocity - initial velocity / time

a = 3.1 m/s - (-6.3 m/s) / 0.02 s = 470 m/s²

The direction of the acceleration is upward, perpendicular to the ground.

The vector average acceleration will be:

a = (0, 470 m/s²) or (470 m/s² * cos 90º, 470 m/s² * sin 90º)

4 0
25 days ago
A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = 0.01t4 − 0.0
serg [3582]
Since you've completed parts a and b, I will tackle part c.
For part C
To respond to this question, we must identify the zeros of the velocity function:
v(t)=0.04t^3-0.06t^2
This polynomial can be factored:
v(t)=0.04t^3-0.06t^2=t^2(0.04t-0.06)
Finding the zeros now becomes straightforward since the function equals zero when any factor is zero.
t^2=0;\\ 0.04t-0.06=0
By solving these equations, we identify our zeros:
t_1=0; t_2=\frac{3}{2}
The particle remains stationary at t=0 and t=3/2.
For part D
We must discover when the velocity function exceeds zero. We will utilize its factored form.
We will assess when each factor is greater than zero and compile the findings in the following table:
\centering \label{my-label} \begin{tabular}{lllll} Range & -\infty & 0 & 3/2 & +\infty \\ t^2 & - & + & + & + \\ 0.04t-0.06 & - & - & + & + \\ t^2 (0.04t-0.06) & + & - & + & + \end{tabular}
From the table, it's evident that our function is positive when - \infty < t and t>3/2.
This indicates the interval during which the particle moves forward.
For part E
The distance traveled can be represented as:
s(t)=0.01t^4 - 0.02t^3
We simply substitute t=12 to calculate the total distance traveled:
s(12)=0.01(12)^4 - 0.02(12)^3=172.80 ft
For part F
Acceleration is defined as the rate at which velocity changes.
We determine acceleration by deriving the velocity function concerning time.
a(t)=\frac{dv}{dt}=(0.04t^3-0.06t^2)'=0.12t^2-0.12t
To find the acceleration at 1 second, we substitute t=1s into the previous equation:
a(1)=0.12-0.12=0


7 0
1 month ago
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