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3 years ago

What is the mass of a dragster with a net force of 27,000N and acceleration of 30.0 m/s^2

Physics

1 answer:

garri49 [273]3 years ago

3 0

Answer:

900kg

Explanation:

Data obtained from the question include:

F = 27000N

a = 30.0m/s^2

M =?

F = Ma

M = F/a

M = 27000/30

M = 900kg

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Draw the ray diagram of Radius of concave mirror=7cm

seropon [69]

Answer:

Nature and size of image:

the image is formed at the focus (F), the image is virtual,erect and highly diminished.

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3 years ago

Can you get your sphere to ever be fully lit (lit on both halves)? If not, why?

BartSMP [9]

Answer:

Explanation:

think of the sphere like earth, light isnt on both sides during day. Its just like a sphere it wont be fully light.

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3 years ago

Read 2 more answers

You blow across the open mouth of an empty test tube and producethe fundamental standing wave of the air column inside the testt

sertanlavr [38]

Answer:

(a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

Explanation:

Given that,

Length of tube = 11.0 cm

(a). We need to calculate the frequency of this standing wave

Using formula of fundamental frequency

$f_{1}=\dfrac{v}{4l}$

Put the value into the formula

$f_{1}=\dfrac{344}{4\times0.11}$

$f_{1}=781.81\ Hz$

$f_{1}=0.782\ kHz$

(b). If the test tube is half filled with water

When the tube is half filled the effective length of the tube is halved

We need to calculate the frequency

Using formula of fundamental frequency of the fundamental standing wave in the air

$f_{1}=\dfrac{v}{4(\dfrac{L}{2})}$

Put the value into the formula

$f_{1}=\dfrac{344}{4\times\dfrac{0.11}{2}}$

$f_{1}=1563.63\ Hz$

$f_{1}=1.563\ kHz$

Hence, (a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

6 0

4 years ago

A ball is dropped from a rooftop 60m high. <br> How long is the ball in the air?

alina1380 [7]

Answer: 3.49 s

Explanation:

We can solve this problem with the following equation of motion:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0 m$ is the final height of the ball

$y_{o}=60 m$ is the initial height of the ball

$V_{o}=0 m/s$ is the initial velocity (the ball was dropped)

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

$t$ is the time

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (2)

$t=\sqrt{\frac{2 (60 m)}{9.8 m/s^{2}}}$ (3)

Finally we find the time the ball is in the air:

$t=3.49 s$ (4)

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4 years ago

What is the difference between the parts of the plot with positive slope and the parts with negative slope?

zalisa [80]

The difference between the parts of the plot with positive slope and the parts with negative slope can be found below.

<h3>What is a slope?</h3>

A slope in a graph is the ratio of the vertical and horizontal distances between two points on a line. Slope shows both steepness and direction of values.

A slope can either be positive or negative depending on the relationship between the variables being plotted.

A positive slope means that the variable are directly related while the negative slope means that two variables are negatively related.

A positive slope further means that the line moves upward when going from left to right on the graph while with the negative slope, the line moves down when going from left to right.

Learn more about slope at: brainly.com/question/2491620

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2 years ago