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3 years ago

14

A man pushes with a force of 300 N on a bookcase of mass 75 kg. The coefficient of static friction between the bookcase and the

floor is 0.9. Because of friction, the bookcase does not move. What is the force of friction on the bookcase?
A. 362 N
B. 300 N
C. 551 N
D. 662 N

2 answers:

ad-work [718]3 years ago

6 0

A! It’s A. A. It’s a. A

kondaur [170]3 years ago

4 0

362 N is the force of friction on the bookcase.

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Two speakers are spaced 15 m apart and are both producing an identical sound wave. You are standing at a spot as pictured. What

IgorLugansk [536]

Answer:

Frequency is $213.04\ s^{-1}.$

Explanation:

Distance between source 1 from the receiver , $S_1 =\sqrt{10^2+22^2}=24.17\ m.$

Distance between source 2 from the receiver , $S_2=\sqrt{5^2+22^2}=22.56\ m.$

Now ,

Path difference , $r = S_1-S_2=24.17-22.56=1.61\ m.$

We know, for constructive interference path difference should be integral multiple of wavelength .

Therefore, $r=n\times \lambda$

It is given that n = 1,

Therefore, $\lambda=1.61\ m.$

Frequency can be found by , $\nu=\dfrac{v}{\lambda}= \dfrac{343}{1.61}= 213.04\ s^{-1} .$

Hence, this is the required solution.

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3 years ago

Pascal in term fundamental unit is?

Aleksandr-060686 [28]

Answer: SI unit of pressure

Explanation: The pascal (pronounced pass-KAL and abbreviated Pa) is the unit of pressure or stress in the International System of Units (SI). Reduced to base units in SI, one pascal is one kilogram per meter per second squared; that is, 1 Pa = 1 kg · m-1 · s-2.

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3 years ago

2200 kg semi truck driving down the highway has lost control. The truck rolls across the median and into oncoming traffic. The t

serious [3.7K]

Answer:

The semi truck travels at an initial speed of 69.545 meters per second downwards.

Explanation:

In this exercise we see a case of an entirely inellastic collision between the semi truck and the car, which can be described by the following equation derived from Principle of Linear Momentum Conservation: (We assume that velocity oriented northwards is positive)

$m_{S}\cdot v_{S}+m_{C}\cdot v_{C} = (m_{S}+m_{C})\cdot v$ (1)

Where:

$m_{S}$ , $m_{C}$ - Masses of the semi truck and the car, measured in kilograms.

$v_{S}$ , $v_{C}$ - Initial velocities of the semi truck and the car, measured in meters per second.

$v$ - Final speed of the system after collision, measured in meters per second.

If we know that $m_{S} = 2200\,kg$ , $m_{C} = 2000\,kg$ , $v_{C} = 45\,\frac{m}{s}$ and $v = -15\,\frac{m}{s}$ , then the initial velocity of the semi truck is:

$m_{S}\cdot v_{S} = (m_{S}+m_{C})\cdot v -m_{C}\cdot v_{C}$

$v_{S} = \frac{(m_{S}+m_{C})\cdot v - m_{C}\cdot v_{C}}{m_{S}}$

$v_{S} = \left(1+\frac{m_{C}}{m_{S}} \right)\cdot v - \frac{m_{C}}{m_{S}} \cdot v_{C}$

$v_{S} = v +\frac{m_{C}}{m_{S}}\cdot (v-v_{C})$

$v_{S} = -15\,\frac{m}{s}+\left(\frac{2000\,kg}{2200\,kg} \right) \cdot \left(-15\,\frac{m}{s}-45\,\frac{m}{s} \right)$

$v_{S} = -69.545\,\frac{m}{s}$

The semi truck travels at an initial speed of 69.545 meters per second downwards.

3 0

2 years ago

A laser is shone through a double slit and a particular interference pattern is observed on a screen some distance away. If the

Grace [21]

If the separation between the openings in a laser is increased, then the distance between the interference fringes decreases

<h3>What is Interference fringe ?</h3>

Interference fringe refers to bands caused by different lights which can be found in phase or not each other.

Distances between laser fringes are short which is due to light wavelength.
The interference fringes can be estimated by knowing slit separation and wavelength.

In conclusion, if the separation between the openings in a laser is increased, then the distance between the interference fringe decreases

Learn more about Interference fringe here:

brainly.com/question/14264436

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5 0

2 years ago

Consider the three dip1acement vectors A = (3i - 3j) m, B = (i-4j) m, and C = (-2i + 5j) m. Use the component method to determin

tia_tia [17]

Answer with Explanation:

We are given that

A=3i-3j m

B=i-4 j m

C=-2i+5j m

a. $D=A+B+C$

$D=3i-3j+i-4j-2i+5j$

$D=2i-2j$

Compare with the vector r=xi+yj

We get x=2 and y=-2

Magnitude= $\mid D\mid=\sqrt{x^2+y^2}=\sqrt{(2)^2+(-2)^2}=2\sqrt 2$ units

By using the formula $\mid r\mid=\sqrt{x^2+y^2}$

Direction: $\theta=tan^{-1}\frac{y}{x}$

By using the formula

Direction of D: $\theta=tan^{-1}(\frac{-2}{2})=tan^{-1}(-1)=tan^{-1}(-tan45^{\circ})=-45^{\circ}$

b.E=-A-B+C

$E=-3i+3j-i+4j-2i+5j$

$E=-6i+12j$

$\mid E\mid=\sqrt{(-6)^2+(12)^2}=13.4$ units

Direction of E= $\theta=tan^{-1}(\frac{12}{-6}=tan^{-1}(-2)=-63.4^{\circ}$

4 0

3 years ago