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3 years ago

Help me I do not understand

Physics

1 answer:

ololo11 [35]3 years ago

4 0

Answer:

170N

Explanation:

First add 530N to 150N and you get 680N, then add 400N to 450N and get 850N. So subtract 850N by 680N and you get 170N

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A tank of volume 0.25 ft is designed to contain 50 standard cubic feet of air. The temperature is 80° F.Calculate the pressure i

slavikrds [6]

Answer:

The inside Pressure of the tank is $4499.12 lb/ft^{2}$

Solution:

As per the question:

Volume of tank, $V = 0.25 ft^{3}$

The capacity of tank, $V' = 50ft^{3}$

Temperature, T' = $80^{\circ}F$ = 299.8 K

Temperature, T = $59^{\circ}F$ = 288.2 K

Now, from the eqn:

PV = nRT (1)

Volume of the gas in the container is constant.

V = V'

Similarly,

P'V' = n'RT' (2)

Also,

The amount of gas is double of the first case in the cylinder then:

n' = 2n

$\]frac{n'}{n} = 2$

where

n and n' are the no. of moles

Now, from eqn (1) and (2):

$\frac{PV}{P'V'} = \frac{nRT}{n'RT'}$

$P' = 2P\frac{T'}{T} = 2\times 2116\times \frac{299.8}{288.2} = 4499.12 lb/ft^{2}$

7 0

3 years ago

A projectile is launched at ground level with an initial speed of 54.5 m/s at an angle of 35.0° above the horizontal. It strikes

Alchen [17]

<h2>Answer: x=125m, y=48.308m</h2>

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which we have two components: x-component and y-component. Being their main equations to find the position as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=54.5m/s$ is the projectile's initial speed

$\theta=35\°$ is the angle

$t=2.80s$ is the time since the projectile is launched until it strikes the target

$x$ is the final horizontal position of the projectile (the value we want to find)

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=0$ is the initial height of the projectile (we are told it was launched at ground level)

$y$ is the final height of the projectile (the value we want to find)

$g=9.8m/s^{2}$ is the acceleration due gravity

Having this clear, let's begin with x (1):

$x=(54.5m/s)cos(35\°)(2.8s)$ (3)

$x=125m$ (4) This is the horizontal final position of the projectile

For y (2):

$y=0+(54.5m/s)sin(35\°)(2.8s)-\frac{(9.8m/s^{2})(2.8s)^{2}}{2}$ (5)

$y=48.308m$ (6) This is the vertical final position of the projectile

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3 years ago

Do you think most people follow the recommended guidelines for an adequate exercise routine?

blondinia [14]

Personally, I know that I don't however it all depends on the person's lifestyle, health, and resoucres, such as if they are able to complete the excersise or have enough money fore equpemnt. To be safe, I would say that the majority of people do follow the guidelines although they may manipulate it to fit their best interest.

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3 years ago

A 61 kg skater is traveling at 2.5 m/s while carrying a 4.0 kg bowling ball. After he throws the bowling ball forward at twice t

gregori [183]

The final velocity of the skater is 2.34 m/s forward

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the system before and after the ball is thrown must be conserved, in absence of external forces.

Before the ball is thrown, the total momentum is:

$p_i = (M+m)u$