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2 years ago

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A 5kg block is pulled across a table by a horizontal force of 40 N with a frictional force of 8 N opposing the motion. Draw a fo

rce diagram and determine the net force. Then calculate the acceleration of the object.

1 answer:

Taya2010 [7]2 years ago

3 0

Answer:

Net force on the block is 32 N.

Acceleration of the object is 6.4 m/s².

Explanation:

Let the acceleration of the object be $a$ m/s².

Given:

Mass of the block is, $m=5\ kg$

Force of pull is, $F=40\ N$

Frictional force on the block is, $f=8\ N$

The free body diagram of the object is shown below.

From the figure, the net force in the forward direction is given as:

$F_{net}=F-f=40-8=32\ N$

Now, from Newton's second law of motion, net force is equal to the product of mass and acceleration. So,

$F_{net}=ma\\32=5a\\a=\frac{32}{5}=6.4\ m/s^2$

Therefore, the acceleration of the object in the forward direction is 6.4 m/s².

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A wheel rotates with a constant angular acceleration of  rad/s2. During a certain time interval its angular displacement is  r

Hatshy [7]

Answer:

The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

Explanation:

Given that,

Angular acceleration $\alpha=\pi\ rad/s^2$

Angular displacement $\theta=\pi\ rad$

Angular velocity $\omega =2\pi\ rad/s$

We need to calculate the angular velocity at the beginning

Using formula of angular velocity

$\alpha =\dfrac{\omega^2-\omega_{0}^2}{2\theta}$

$\omega_{0}^2=\omega^2-2\alpha\theta$

Where, $\alpha$ = angular acceleration

$\omega$ = angular velocity

Put the value into the formula

$\omega_{0}^2=(2\pi)^2-2\times\pi\times\pi$

$\omega=\sqrt{2\pi^2}$

$\omega_{0}=\pi\sqrt{2}\ rad/s$

Hence, The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

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2 years ago

A very tall building has a height H0 on a cool spring day when the temperature is T0. You decide to use the building as a sort o

Vlada [557]

Answer:

The temperature is $T = \frac{h}{H_O \alpha_{steel} } + T_O$

Explanation:

From the question we are told that

The height on a cool spring day is $H_O$

The temperature on a cool spring day is $T_O$

The difference in height between a cool spring day and a summer day is h

The coefficient of static friction is $\alpha _{steel}$

The mathematical relation for the linear expansion of the steel buiding is represented as

$h = H_o \alpha_{steel} [T-T_O]$

Where T is the temperature of the steel during summer

Now making T the subject we have

$T = \frac{h}{H_O \alpha_{steel} } + T_O$

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2 years ago

Kendra wants to use similes to describe the reflection and absorption of waves. Which pair of similes best fit?

Scrat [10]

Answer:

Reflection is like bouncing a tennis ball, and absorption is like water soaking into a paper towel.

Reflection is like water soaking into a paper towel, and absorption is like bouncing a tennis ball.

Reflection is the way a straw appears in a glass of water, and absorption is the separation of colors from a prism.

Reflection is the separation of colors from a prism, and absorption is the way a straw appears in a glass of water

Explanation:

already asked

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2 years ago

In a certain clock, a pendulum of length L1 has a period T1 = 0.95s. The length of the pendulum

gulaghasi [49]

Answer:

Ratio of length will be $\frac{L_2}{L_1}=1.108$

Explanation:

We have given time period of the pendulum when length is $L_1$ is $T_1=0.95sec$

And when length is $L_2$ time period $T_2=1sec$

We know that time period is given by

$T=2\pi \sqrt{\frac{L}{g}}$

So $0.95=2\pi \sqrt{\frac{L_1}{g}}$ ----eqn 1

And $1=2\pi \sqrt{\frac{L_2}{g}}$ -------eqn 2

Dividing eqn 2 by eqn 1

$\frac{1}{0.95}=\sqrt{\frac{L_2}{L_1}}$

Squaring both side

$\frac{L_2}{L_1}=1.108$

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2 years ago

The ends of a massless rope are attached to two stationary objects (e.g., two trees or two cars) so that the rope makes a straig

julia-pushkina [17]

Answer: They are all true

a. The tension in the rope is everywhere the same.

b. The magnitudes of the forces exerted on the two objects by the rope are the same.

c. The forces exerted on the two objects by the rope must be in opposite directions.

d. The forces exerted on the two objects by the rope must be in the direction of the rope.

Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great rest of Black History Month! :-)

- Cutiepatutie ☺❀❤

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2 years ago