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2 years ago

11

A student sees a newspaper ad for an apartment that has 2030 square feet (ft2) of floor space. How many square meters of area ar

e there?

1 answer:

Vilka [71]2 years ago

8 0

Square feet or square meters?

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If a 7 kg bowling ball is lifted 2 m into the air and dropped, what speed will it strike the ground? (

Naily [24]

Answer:

6.32m/s

Explanation:

note:Now these calculations are based in the fact that acc. due to gravity is 10m/s²

okay so I'm thinking you think the speed of a body depends on the mass of the body also,umh... well it doesn't at all!

when two bodies of different masses fall from the same height,they fall at the same time( this is just to say)

now enough of the talking let solve....

so the ball was dropped .ie from rest to the ground through a distance of 2m,

the formula for calculating the distance if a body moving in a straight line is given by:

S=ut + ½at² where u is initial velocity, a is acceleration ( of the body or due to gravity, but since its falling freely under the influence of gravity its " we use the acceleration due to gravity ,which is 10m/s²) and t is the time taken to cover the distance.

from our question the ball was dropped from rest thus its u is 0 therefore we use this equation to find the time it took to touch ground (S=½at²)

solving ....

we get t to be 0.632s

to find the speed we substitute t in the equation below:

V=u+at ,but since u=0

V=at =10•0.632=6.32m/s

therefore the speed the body uses to strike the ground is 6.32m/s

4 0

2 years ago

The net horizontal force on a car is 981 N. The car has a mass of 1550 kg and the force is applied when the car has a speed of 2

viktelen [127]

Answer:

Distance, d = 778.05 m

Explanation:

Given that,

Force acting on the car, F = 981 N

Mass of the car, m = 1550 kg

Initial speed of the car, v = 25 mi/h = 11.17 m/s

We need to find the distance covered by car if the force continues to be applied to the car. Firstly, lets find the acceleration of the car:

$F=ma\\\\a=\dfrac{F}{m}\\\\a=\dfrac{981}{1550}\\\\a=0.632\ m/s^2$

Let d is the distance covered by car. Using second equation of motion as :

$d=ut+\dfrac{1}{2}at^2\\\\d=11.17\times 35+\dfrac{1}{2}\times 0.632\times (35)^2\\\\d=778.05\ m$

So, the car will cover a distance of 778.05 meters.

5 0

3 years ago

An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get

Blizzard [7]

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

$mV_i = (m-m_O)V_f - m_OV_O$

where,

m=Total mass

$m_O =$ Mass of Object

$V_i =$ Velocity before throwing

$V_f =$ Final Velocity

$V_O =$ Velocity of Object

Our values are:

$m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg$

Solving to find the final speed, after throwing the object we have

$V_f=\frac{mV_0+m_TV_O}{m-m_O}$

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) $5.3Kg\rightarrow 15m/s$

$V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}$

$V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}$

$V_{f1}=0.8511m/s$

B) $7.9Kg\rightarrow 11.2m/s$

$V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}$

$V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}$

$V_{f2} = 2.0173m/s$

C) $10.5Kg\rightarrow 7m/s$

$V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}$

$V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}$

$V_{f3} = 3.63478m/s$

Therefore the final velocity of astronaut is 3.63m/s

7 0

3 years ago

Two observers in different inertial reference frames moving relative to each other at nearly the speed of light see the same two

lions [1.4K]

Answer:

The correct answer is d Both the observer's are correct

Explanation:

We know by postulates of relativity that laws of physics are same in different inertial frames.

Thus for each of the frames they make observations related to their frames and since the observations are true for their individual frames they both are correct. But when we compare the two frames we need to use transformation equations to compare both the results.

3 0

3 years ago

This problem has been solved!

lana66690 [7]

Answer:

Charge on each metal sphere will be $8\times 10^{8}C$

Explanation:

We have given number of electron added to metal sphere A $n=10^{12}electron$

As both the spheres are connected by rod so half -half electron will be distributed on both the spheres.

So electron on both the spheres $=\frac{10^{12}}{2}=5\times 10^{11}electron$

We know that charge on each electron $e=1.6\times 10^{-19}C$

So charge on both the spheres will be equal to $q=1.6\times 10^{-19}\times 5\times 10^{11}=8\times 10^{8}C$

So charge on each metal sphere will be equal to $8\times 10^{8}C$

6 0

2 years ago