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3 years ago

SOMEONE HELP! If I don't get this done i will fail, I have been trying for hours.

Physics

1 answer:

uysha [10]3 years ago

4 0

Answer:

1. The baseball will hit the bat at a force of 1.875 Newtons (N)

2. The basketball is being pushed to roll to the right of its original position with an applied rightward force. There is a force of friction to the left, a normal force of the floor acting upwards on the ball, and a gravity force of the Earth acting downwards on the ball.

Explanation:

1. To calculate force, you just have to multiply mass (m) and acceleration (a). F = m * a --> F = 0.075kg * 25 m/s^2 = 1.875 N.

2. Use your knowledge of different types of forces and free-body diagrams to come to this conclusion.

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Four identical capacitors are connected with a resistor in two different ways. When they are connected as in part a of the drawi

Nina [5.8K]

Answer:

$T_2 = 0.592$

Explanation:

Given

$T_1 = 1.48s$

See attachment for connection

Required

Determine the time constant in (b)

First, we calculate the total capacitance (C1) in (a):

The upper two connections are connected serially:

So, we have:

$\frac{1}{C_{up}} = \frac{1}{C} + \frac{1}{C}$

Take LCM

$\frac{1}{C_{up}} = \frac{1+1}{C}$

$\frac{1}{C_{up}}= \frac{2}{C}$

Cross Multiply

$C_{up} * 2 = C * 1$

$C_{up} * 2 = C$

Make $C_{up}$ the subject

$C_{up} = \frac{1}{2}C$

The bottom two are also connected serially.

In other words, the upper and the bottom have the same capacitance.

So, the total (C) is:

$C_1 = 2 * C_{up}$

$C_1 = 2 * \frac{1}{2}C$

$C_1 = C$

The total capacitance in (b) is calculated as:

First, we calculate the parallel capacitance (Cp) is:

$C_p = C+C$

$C_p = 2C$

So, the total capacitance (C2) is:

$\frac{1}{C_2} = \frac{1}{C_p} + \frac{1}{C} + \frac{1}{C}$

$\frac{1}{C_2} = \frac{1}{2C} + \frac{1}{C} + \frac{1}{C}$

Take LCM

$\frac{1}{C_2} = \frac{1 + 2 + 2}{2C}$

$\frac{1}{C_2} = \frac{5}{2C}$

Inverse both sides

$C_2 = \frac{2}{5}C$

Both (a) and (b) have the same resistance.

So:

We have:

Time constant is directional proportional to capacitance:

So:

$T\ \alpha\ C$

Convert to equation

$T\ =kC$

Make k the subject

$k = \frac{T}{C}$

$k = \frac{T_1}{C_1} = \frac{T_2}{C_2}$

$\frac{T_1}{C_1} = \frac{T_2}{C_2}$

Make T2 the subject

$T_2 = \frac{T_1 * C_2}{C_1}$

Substitute values for T1, C1 and C2

$T_2 = \frac{1.48 * \frac{2}{5}C}{C}$

$T_2 = \frac{1.48 * \frac{2}{5}}{1}$

$T_2 = \frac{0.592}{1}$

$T_2 = 0.592$

Hence, the time constance of (b) is 0.592 s

8 0

3 years ago

Because the time was the same for each segment you know the speed was the same for each segment

netineya [11]

Answer:

Market segmentation helps you send the right message, every time, by efficiently targeting specific groups of consumers.

Explanation:

3 0

2 years ago

A racehorse is running with a uniform speed of 69 km/hr along a straightaway. what is the time it takes for the horse to cover 4

Alex

Hello there,
400 meters= 0.4 km
Time= Distance / speed
= 0.4 / 69
= 0.0057971014492754 hr
= 0.35 min

Hope this helps :))

~Top

8 0

3 years ago

Read 2 more answers

at which plate boundary would you expect to see volcanoes A.divergent b.all of the above c.convergent D.transform

natulia [17]

A or d would be most likely it

7 0

4 years ago

If the voltage across a circuit of constant resistance is doubled, the power dissipated by that circuit will

densk [106]

Answer:

The voltage will quadruple

Explanation:

The power dissipated in a circuit is given by

$P=\frac{V^2}{R}$

where

V is the voltage

R is the resistance

In this problem, the voltage across the circuit is doubled:

V' = 2V

So the new power dissipated is

$P'=\frac{V'^2}{R}=\frac{(2V)^2}{R}=4\frac{V^2}{R}=4 P$

so, the power dissipated will quadruple.

6 0

4 years ago