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9 days ago

13

A baseball player throws a ball straight up and catches it 3.0 s later. If the ball accelerates at 9.85 m/s2 down, at what veloc

ity did he throw the ball?

1 answer:

Softa [2K]9 days ago

7 0

Bajo una aceleración constante, se establece que al final del lanzamiento, la velocidad de la pelota es equivalente a su velocidad inicial en magnitud, pero se presenta en dirección descendente, para lo cual su signo es opuesto.

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A student pulls a block over a rough surface with a constant force FP that is at an angle θ above the horizontal, as shown above

serg [2593]

Response:

B. The friction force between the block and surface will diminish.

Clarification:

The force of friction can be expressed as

$F_s = \mu N$

where $\mu$ represents the friction coefficient and $N$ signifies the normal force.

As the student pulls the block with force $F_p$ at an angle $\theta$ , the normal force acting on the block subsequently becomes

$N = Mg- F_psin(\theta)$

and therefore, the frictional force changes to

$F_s = \mu (Mg- F_psin(\theta))$ .

With an increase in $\theta$ , the $sin(\theta)$ increases, which subsequently reduces the normal force $Mg- F_psin(\theta)$ , indicating a decrease in the frictional force; hence choice B is correct.

Note: While choice D seems appealing, it is incorrect since the weight $W=mg$ remains unaffected by the external forces acting on the block.

6 0

25 days ago

Read 2 more answers

A book slides across a level, carpeted floor with an initial speed of 3.25 m/s and comes to rest after 3.25 m. Calculate the coe

Maru [2360]

Answer:

The coefficient of kinetic friction calculates to 0.165.

Explanation:

Given that,

Initial speed = 3.25 m/s

Distance traveled = 3.25 m

We must figure out the acceleration

Applying the third equation of motion

$v^2=u^2+2as$

$a=\dfrac{v^2-u^2}{2s}$

Where a = acceleration

u = starting velocity

v = ending velocity

s = distance

Insert the values into the formula

$a=\dfrac{-(3.25)^2}{2\times3.25}$

$a=-1.625\ m/s^2$

The negative sign indicates deceleration

Next, we calculate the coefficient of kinetic friction.

Using the frictional force

$F_{k} =k mg$ ....(I)

Where k = kinetic friction coefficient

m = mass

g = gravitational acceleration

Utilizing Newton's second law

$F = ma$ ....(II)

Setting equations (I) and (II) equal to each other

$k mg=ma$

$k=\dfrac{a}{g}$

$k=\dfrac{1.625}{9.8}$

$k=0.165$

Thus, the coefficient of kinetic friction is determined to be 0.165.

8 0

16 days ago

A 2600-m-high mountain is located on the equator. how much faster does a climber on top of the mountain move than a surfer at a

Maru [2360]

The climber's speed is 0.19 m/s greater than that of the surfer on the beach.

Both individuals are on Earth and share a consistent angular velocity, but their linear speeds differ.

Calculating the seconds in a day gives us t=24*60*60=86400 sec

The linear speed on the beach is computed as

V1= $\frac{2πr}{t}$

Where t is the duration

Substituting values into the equation leads to

V1= $\frac{2π*6.4*10^6}{86400}$ =465.421 m/s

The mountain's linear speed is determined as

V2= $\frac{2π(r+h)}{t}$

Inserting values into the equation results in

V2= $\frac{2π(6.4*10^6+2600}{86400}$ =465.61 m/s

Thus, the climber exceeding the surfer's speed is calculated as 465.61-465.421=0.19 m/s

5 0

1 month ago

Solenoid 2 has twice the diameter, twice the length, and twice as many turns as solenoid 1. How does the field B2 at the center

Sav [2230]

Complete Question

The entire question is displayed in the first uploaded image

Answer:

The right choice is option 3

Explanation:

The question informs us that

The diameter of solenoid 1 is $d_1$

The length of solenoid 1 is $L_1$

The number of turns of solenoid is $N_1$

The diameter of solenoid 2 is $d_2 = 2d_1$

The length of solenoid 2 is $L_2 = 2L_1$

The number of turns of solenoid 2 is $N_2 = 2 N_1$

Typically, the magnetic field in a solenoid can be expressed mathematically as

$B = \frac{\mu_o * N * I }{L}$

According to this formula we find that

$B \ \alpha \ \frac{N}{L}$

$B = C \frac{N}{L}$

Here C denotes constant

=> $C = \frac{B * \frac{L}{N}$

=> $\frac{B_1 * \frac{L_1}{N_1} = \frac{B_2 * \frac{L_2}{N_2}$

=> $\frac{B_1}{B_2 } = \frac{N_1 L _2}{ N_2L_1}$

=> $\frac{B_1}{B_2 } = \frac{N_1 * (2 L_1)}{ (2 N_2)L_1}$

=> $\frac{B_1}{B_2 } = 1$

=> $B_2 = B_1$

4 0

14 days ago

Mr. Chang's class participated in a science experiment where they placed three substances outside in the sunlight to determine w

Sav [2230]

It could be B or D, but I am fairly certain it is D.

7 0

3 days ago