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4 years ago

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A 4 kg block is pulled with 5.6 N of force to the right. The block experiences 1.1 N of friction. What is the acceleration of th

e block?

1 answer:

BARSIC [14]4 years ago

7 0

Answer:

The acceleration of the box is 1.125 m/s² towards right.

Explanation:

Mass of the box, $m=4$ kg

Force acting towards right, $F=5.6$ N

Frictional force acting towards left, $f=1.1$ N

Let the acceleration be $a$ m/s².

Now, net force acting on the box towards right is given as:

$F_{net}=F-f=5.6-1.1=4.5\textrm{ N}$

From Newton's second law of motion,

$F_{net}=ma\\4.5=4a\\a=\frac{4.5}{4}=1.125\textrm{ }m/s^2$

Therefore, the acceleration of the box is 1.125 m/s² towards right.

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Answer:

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Explanation:

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3 years ago

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3 years ago

Human centrifuges are used to train military pilots and astronauts in preparation for high-g maneuvers. A trained, fit person we

jeka57 [31]

(a) $4.14 rad/s^2$

The relationship beween centripetal acceleration and angular speed is

$a=\omega^2 r$

where

$\omega$ is the angular speed

r is the radius of the circular path

Here we gave

$a = 9g = 88.2 m/s^2$ is the centripetal acceleration

r = 5.15 m is the radius

Solving for $\omega$ , we find:

$\omega = \sqrt{\frac{a}{r}}=\sqrt{\frac{88.2 m/s^2}{5.15 m}}=4.14 rad/s^2$

(b) 21.3 m/s

The relationship between the linear speed and the angular speed is

$v=\omega r$

where

v is the linear speed

$\omega$ is the angular speed

r is the radius of the circular path

In this problem we have

$\omega=4.14 rad/s$

r = 5.15 m

Solving the equation for v, we find

$v=(4.14 rad/s)(5.15 m)=21.3 m/s$

7 0

3 years ago

Read 2 more answers

Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV

Korolek [52]

Complete Question

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).

Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius R which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm2.

How large does the radius R of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver?

For simplicity, assume that your house is located directly beneath the satellite (i.e. the situation you calculated in the first part), that the dish reflects all of the incident signal onto the receiver, and that there are no losses associated with the reception process. The dish has a curvature, but the radius R refers to the projection of the dish into the plane perpendicular to the direction of the incoming signal.

Give your answer in centimeters, to two significant figures.

Answer:

The radius of the dish is $R = 18cm$

Explanation:

From the question we are told that

The radius of the orbit is = $R = 35,000km = 35,000 *10^3 m$

The power output of the power is $P = 1 kW = 1000W$

The electric vector amplitude is given as $E = 0.1 mV/m = 0.1 *10^{-3}V/m$

The area of thereciever is $A_R = 5cm^2$

Generally the intensity of the dish is mathematically represented as

$I = \frac{P}{A}$

Where A is the area orbit which is a sphere so this is obtained as

$A = 4 \pi r^2$

$= (4 * 3.142 * (35,000 *10^3)^2)$

$=1.5395*10^{16} m^2$

Then substituting into the equation for intensity

$I_s = \frac{1000}{1.5395*10^{16}}$

$= 6.5*10^ {-14}W/m2$

Now the intensity received by the dish can be mathematically evaluated as

$I_d = \frac{1}{2} * c \epsilon_o E_D ^2$

Where c is thesped of light with a constant value $c = 3.0*10^8 m/s$

$\epsilon_o$ is the permitivity of free space with a value $8.85*10^{-12} N/m$

$E_D$ is the electric filed on the dish

So since we are to assume to loss then the intensity of the satellite is equal to the intensity incident on the receiver dish

Now making the eletric field intensity the subject of the formula

$E_D = \sqrt{\frac{2 * I_d}{c * \epsilon_o} }$

substituting values

$E_D = \sqrt{\frac{2 * 6.5*10^{-14}}{3.0*10^{8} * 8.85*10^{-12}} }$

$= 7*10^{-6} V/m$

The incident power on the dish is what is been reflected to the receiver

$P_D = P_R$

Where $P_D$ is the power incident on the dish which is mathematically represented as

$P_D = I_d A_d$

$= \frac{1}{2} c \epsilon_o E_D^2 (\pi R^2)$

And $P_R$ is the power incident on the dish which is mathematically represented as

$P_R = I_R A_R$

$= \frac{1}{2} c \epsilon_o E_R^2 A_R$

Now equating the two

$\frac{1}{2} c \epsilon_o E_D^2 (\pi R^2) = \frac{1}{2} c \epsilon_o E_R^2 A_R$

Making R the subject we have

$R = \sqrt{\frac{E_R^2 A_R}{\pi E_D^2} }$

Substituting values

$R = \sqrt{\frac{(0.1 *10^{-3})^2 * 5}{\pi (7*10^{-6})^ 2} }$

$R = 18cm$

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3 years ago

Which of the following best illustrates a pair of sentences that are joined by an understood relationship?

zubka84 [21]

Answer : Option C) Detective Smiley scanned the dim hallway. He pulled his pistol from its holster.

Explanation : Amongst the given other choices there seems to be no relationship between two consecutive sentences. The only sentence which seemed to have a connection between previous and later sentence was option C. Where it is clearly stated that the detective named as Smiley was walking through the hallway which was dimly lit. The second sentence has a co-relation to the previous one as it extends the sentence. Detective smiley then pulled out his pistol from his holster after walking through the hallway.

This confirms that the correct answer for this question is Option C.

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3 years ago

Read 2 more answers