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3 years ago

I need this ASAP please!!!

Physics

2 answers:

WINSTONCH [101]3 years ago

4 0

It’s E we just had a test in this and I got it right

Mademuasel [1]3 years ago

3 0

Answer:

C. the help our understanding of the formation of planetary bodies.

Explanation:

this is the answer of PLATO

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A different force is applied to each of four different blocks on a frictionless, horizontal surface. In which diagram does the b

NeTakaya

Answer:

C. 4

Explanation:

6 0

3 years ago

Vector C has a magnitude of 24.6 m and points in the − y ‑ direction. Vectors A and B both have positive y ‑ components, and mak

frez [133]

Answer:

A= 61.35

B= -44.40

Explanation:

1. Using the components method we have:

$A_{x}= A cos \alpha\\B_{x}= B cos \beta\\C_{x}= 0\\\\A_{y}= A sin \alpha\\B_{y}= B sin \beta\\C_{y}= 24.6\\$

Considering that the vector sum $A+B+C=0$ , then:

$|V|=\sqrt{V_{x}^{2} +V_{y}^{2} }=0$

Then:

$V_{x} ^{2} =0; V_{x} =0\\V_{y} ^{2} =0; V_{y} =0$

It means the value of x and y component is 0.

2. Determinate the equations that describe each component:

$V_{x}= A cos \alpha -B cos \beta=0 (1)\\V_{y}= A sin \alpha +B sin \beta - C=0 (2)$

Form Eq. (1):

$A=B \frac{cos \beta}{cos \alpha} (3)$

Replacing A in Eq. (2):

$(B\frac{cos \beta}{cos \alpha})(sin \alpha)+ B sin\beta-C=0\\(B\frac{cos \beta}{cos \alpha})(sin \alpha)+ B sin\beta-=C\\\\B(\frac{cos \beta. sin \alpha}{cos \alpha}+ sin\beta)=C\\B=C(\frac{cos \beta. sin \alpha}{cos \alpha}+ sin\beta)^{-1} (4)$

Replacing values of C, α and β in (4):

$B= 24.6 (\frac{(cos 27.7)(sin 44.9)}{cos 44.9}+sin 27.7)^{-1} \\B= -44.4$

Replacing value of B in (3)

$A=-44.40\frac{cos 27.7}{sin 49.9} \\A= 61.35$

5 0

3 years ago

A rocket moves upward, starting from rest with an acceleration of 25.4 m/s^2 for 3.39 s. It runs out of fuel at the end of the 3

kolbaska11 [484]

Explanation:

Initial speed of the rocket, u = 0

Acceleration of the rocket, $a=25.4\ m/s^2$

Time taken, t = 3.39 s

Let v is the final velocity of the rocket when it runs out of fuels. Using the equation of kinematics as :

$v=u+at$

$v=25.4\times 3.39=86.10\ m/s$

Let x is the initial position of the rocket. Using third equation of kinematics as :

$v^2=u^2+2ax_o$

$x_o=\dfrac{v^2}{2a}$

$x_o=\dfrac{86.10^2}{2\times 25.4}=145.92\ m$

Let $x_o$ is the position at the maximum height. Again using equation of motion as :

$v^2-u^2=2a(x-x_o)$

Now $a=-g$ and v and u will interchange

$u^2=2g(x-x_o)$

$x=x_o+\dfrac{u^2}{2g}$

$x=145.92+\dfrac{(86.10)^2}{2\times 9.8}$

x = 524.14 meters

Hence, this is the required solution.

5 0

3 years ago

A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 ∘ above t

goldfiish [28.3K]

Answer:

Explanation:

initial height, yo = 2 m

initial velocity, u = 20 m/s

angle of projection,θ = 5 degree

distance of net = 7 m

height of net = 1 m

Let it covers a vertical distance y in time t .

Use Second equation of motion for vertical motion

$y=y_{0}+uSin\theta t-1/2 gt^{2}$

$y=2+20Sin5 t-4.9t^{2}$

As it hits the ground in time t, so put y = 0

$0=2+1.74 t-4.9t^{2}$

$4.9t^{2}-1.74t-2=0$

$t= \frac{1.74\pm\sqrt{1.74^{2}+4\times\2\times4.9}}{9.8}$

Taking positive sign, t = 0.84 s

The ball travels a horizontal distance x in time t

X = 20 Cos5 x t

X = 16.76 m

As this distance is more than the distance of net, so it clears the net.

Let t' be the time taken to travel a horizontal distance equal to the distance of net

7 = 20 cos5 x t'

t' = 0.35 s

Let the vertical distance traveled by the ball in time t' is y'.

So,

$y'=y_{0}+uSin\theta t'-1/2 gt'^{2}$

$y'=2+20Sin5 t-4.9\times0.35^{2}$

y' = 2.008 m

So, it clears the net which is 1 m high.

It clears the net by a vertical distance of 2.008 - 1 = 1.008 m and horizontal distance 16.76 - 7 = 9.76 m

3 0

3 years ago

Negative charges can move from one insulator to another.<br> True<br> False

Rasek [7]

Answer:

True

Explanation:

Because in atom the negative charge become lose or gain.

6 0

3 years ago