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3 years ago

Hi please may someone help me especially on the sketch part.

1 answer:

7 0

Ignoring the air resistance it will take about 3 seconds for the object to reach the ground.We know that the acceleration due to gravity is 10m/s2.

We also know that the final velocity is 30 m/s while the initial velocity is 0 m/s

we can use the formulae for acceleration to calculate the time taken/

(final - initial velocity)/timetaken=10

(30-0)/timetaken=10

timetaken =30/10=3 seconds

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A 39.5 kg child stands at the center of a 125 kg playground merry-go-round which rotates at 3.10 rad/s. If the child moves to th

Neporo4naja [7]

Using the law of conservation of angular momentum, we have

I1 w1 = I2 w2 

ie., m1r^2/2 x w1 = ( m1r^2/2 + m2r^2 ) w2 

ie., new angular velocity w2 = m1 w1 / ( m1+ 2m2) = 125 x 3.1 / ( 125 + 2 x39.5 ) 

= 1.8995 = 1.9 rad /sec ( nearly )

3 0

3 years ago

Billy drops a ball from a height of 1 m. The ball bounces back to a height of 0.8 m, then

Andreas93 [3]

Answer:

The displacement is $\Delta H = - 1 \ m$

The distance is $D = 4 \ m$

Explanation:

From the question we are told that

The height from which the ball is dropped is $h = 1 \ m$

The height attained at the first bounce is $h_1 = 0.8 \ m$

The height attained at the second bounce is $h_2 = 0.5 \ m$

The height attained at the third bounce is $h_3 = 0.2 \ m$

Note : When calculating displacement we consider the direction of motion

Generally given that upward is positive the total displacement of the ball is mathematically represented as

$\Delta H = (0 - h ) + ( h_1 - h_1 ) + (h_2 - h_2 )+ (h_3 - h_3)$

Here the 0 show that there was no bounce back to the point where Billy released the ball

$\Delta H = (0 - 1 ) + ( 0.8- 0.8 ) + (0.5 - 0.5 )+ (0.2 - 0.2)$

=> $\Delta H = - 1 \ m$

Generally the distance covered by the ball is mathematically represented as

$D = h + 2h_2 + 2h_3 + 2h_3$

The 2 shows that the ball traveled the height two times

$D = 1 + 2* 0.8 + 2* 0.5 + 2* 0.2$

=> $D = 4 \ m$

5 0

3 years ago

The water behind Grand Coulee Dam is 1000 m wide and 200 m deep. Find the hydrostatic force on the back of the dam. (Hint: the t

Vika [28.1K]

Answer:

The hydro static force on the back of the dam is $1.96\times10^{11}\ N$

Explanation:

Given that,

Width b= 1000 m

Depth d= 200 m

We need to calculate the average pressure

Using formula of average pressure

$P_{avg}=\rho\times g\times d_{avg}$

Put the value into the formula

$P_{avg}=1000\times9.8\times100$

$P_{avg}=980000\ Pa$

We need to calculate the hydro static force on the back of the dam

Using formula of force

$F = P_{avg}\times A$

Put the value into the formula

$F = 980000\times1000\times200$

$F=1.96\times10^{11}\ N$

Hence, The hydro static force on the back of the dam is $1.96\times10^{11}\ N$

7 0

3 years ago

A makes an angel of 30.0 above the positive x-axis, and B makes an angel of 45.0 below the negative x-axis. A=3.00 units and B=2

erik [133]

I think this is vectors. Sketch the two vectors A and B on the x axis and then find their magnitudes using cosine... i would like to know if i am correct.

5 0

3 years ago

Can an object contain heat? Why or why not?

Stells [14]

Every object has thermal energy (better word than heat, since we associate that with high temperatures). This is actually the molecules vibrating, moving a lot. More thermal energy means more vibrating, and thus also expanding in volume.

8 0

3 years ago