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3 years ago

2. How long will it take for a ball thrown vertically upward with an initial velocity of

Physics

1 answer:

KengaRu [80]3 years ago

6 0

Answer:

13.33 seconds

Explanation:

At maximum height, the equation of motion becomes:

v = u + at

Since the object was thrown vertically, the initial velocity (u) is zero and the acceleration (a) becomes the acceleration due to gravity (10 m/s2). The equation becomes:

v = at

v = 480 km/hr = 133.333 m/s

10t = 133.333

t = 133.333/10

t = 13.33 seconds.

The time for the ball thrown vertically with a velocity of 480 km/hr to reach the maximum height is 13.33 seconds.

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Help pls ASAP thank you

tino4ka555 [31]

Answer: 3rd option

Explanation: Since it has less mass and same force as other object it will have higher acceleration

7 0

3 years ago

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What is the current in a wire of radius R = 2.02 mm if the magnitude of the current density is given by (a) Ja = J0r/R and (b) J

Sloan [31]

Explanation:

For this problem we have to take into account the expression

J = I/area = I/(π*r^(2))

By taking I we have

I = π*r^(2)*J

(a)

For Ja = J0r/R the current is not constant in the wire. Hence

$I(r) = \pi r^{2} J(r) = \pi r^{2} J_{0}r/R = \pi r^{3} (3.74*10^{4}A/m^{2})/(2.02*10^{-3}m)$

and on the surface the current is

$I(R) = \pi r^{2} J(R) = \pi r^{2} J_{0}R/R = \pi(2.02*10^{-3})^{2} (3.74*10^{4}) = 0.47 A$

(b)

For Jb = J0(1 - r/R)

$I(r)=\pi r^{2}J(r) =\pi r^{2} J_{0}(1 - r/R)=\pi r^{2}J_{0}(1-\frac{r}{2.02*10^{-3}} )$

and on the surface

$I(R)=\pi r^{2}J_{0}(1-R/R)=\pi r^{2}J_{0}(1-1)= 0$

(c)

Ja maximizes the current density near the wire's surface

Additional point

The total current in the wire is obtained by integrating

$I_{T}=\pi\int\limits^R_0 {r^{2}Ja(r)} \, dr = \pi \frac{J_{0}}{R}\int\limits^R_0 {r^{3}} \ dr =\pi \frac{J_{0}R^{4}}{4R}=\frac{1}{4}\pi J_{0}R^{3}=2.42*10^{-4} A$

and in a simmilar way for Jb

$I_{T}=\pi J_{0} \int\limits^R_0 {r^{2}(1-r/R)} \, dr = \pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2R}]=\pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2}]$

And it is only necessary to replace J0 and R.

I hope this is useful for you

regards

7 0

3 years ago

The Moon has too little iron, Mercury too much. How can both of these anomalies be the result of giant impacts? Explain how the

arsen [322]

Answer:

Explanation:

Giant impact has resulted into formation of celestial bodies such as moon.

Moon was formed years after the Earth was formed. When Earth collided with very big size bodies, there was vaporization of chunks of the planets in to the space. Gravity plays its role and bounded those particles together thus forming moon.

This formation explains Moon is made up of lighter elements and contains too little iron where as Mercury was formed with the rest of the planets and just like earth, its core is formed first which is of iron and then collected the lighter elements to form crust and mantle.

Though both Moon and Mercury were the result of giant impacts yet they have different elements because mercury was made earlier and its is closer to moon hence it collected heavy elements such as iron to for the formation where as moon is the result of collision of Earth and other Mars- sized body hence its constituent particles are different that of mercury.

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4 years ago

Durning which type of process does pressure remain consistent

Simora [160]

Answer:

$\fbox {D. Isobaric}$

Explanation:

The process during which pressure remains constant is called an isobaric process.

6 0

3 years ago

Read 2 more answers

A sprinter starts from rest and accelerates to her maximum speed of 10.5 m/s in a distance of 11.0 m. (a) What was her accelerat

Anvisha [2.4K]

Answer:

a)a=5.01m/s^2

b)t=11.26s

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t (1)

{Vf^{2}-Vo^2}/{2.a} =X(2)

X=Xo+ VoT+0.5at^{2} (3)

X=(Vf+Vo)T/2 (4)

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

to solve the question a, we can use the ecuation number 2

Vo=0

Vf=10.5 m/s

x=11m

{Vf^{2}-Vo^2}/{2.a} =X

{Vf^{2}-Vo^2}/{2.x} =a

{10.5^{2}-0^2}/{2x11} =a

a=5.01m/s^2

to find the time we can use the ecuation number 1

Vf=Vo+a.t

t=(Vf-Vo)/a

t=(10.5-0)/5.01=2.09s

part b

in this case the spees is constant, so the movement is defined by the following ecuation

X=VT

t=x/v

t=96.3/10.5=9.17s

to find the total time we sum the times when the speed is constant and when the acceleration is constan

t=9.17+2.09

t=11.26s

8 0

3 years ago