Ultraviolet rays from the sun are able to reach Earth's surface because-

An object is thrown vertically up and attains an upward velocity of 9.6 m/s when it reaches one fourth of its maximum height abo

ki77a [65]

Answer:

11.09 m/s

Explanation:

Given that an object is thrown vertically up and attains an upward velocity of 9.6 m/s when it reaches one fourth of its maximum height above its launch point.

The parameters given are:

Initial velocity U = ?

Final velocity V = 9.6 m/s

Acceleration due to gravity g = 9.8m/s^2

Let first assume that the object is thrown from rest with the velocity U, at maximum height V = 0

Using third equation of motion

V^2 = U^2 - 2gH

0 = U^2 - 2 × 9.8H

U^2 = 19.6H ........ (1)

Using the formula again for one fourth of its maximum height

9.6^2 = U^2 - 2 × 9.8 × H/4

92.16 = U^2 - 19.6/4H

92.16 = U^2 - 4.9H

U^2 = 92.16 + 4.9H ...... (2)

Substitute U^2 in equation (1) into equation (2)

19.6H = 92.16 + 4.9H

Collect the like terms

19.6H - 4.9H = 92.16

14.7H = 92.16

H = 92.16/14.7

H = 6.269

Substitute H into equation 2

U^2 = 92.16 + 4.9( 6.269)

U^2 = 92.16 + 30.72

U^2 = 122.88

U = 11.09 m/s

Therefore, the initial velocity of the object is 11.09 m/s

3 0

3 years ago

Estimate the kinetic energy (in GJ) of a 93,000 metric ton aircraft carrier moving at a speed of at 32 knots. You will need to l

Wewaii [24]

Answer:

kinetic Energy = 12.58 GJ

Explanation:

1 metric ton is equal to 1000 kg

then,

93000 metric ton

Mass is m = 93000 x 1000 kg

speed is v = 32 knots

1 knot is 0.514 m/s

then ,

v = 32 x 0.514 = 16.448 m/s

To solve for the Kinetic Energy (KE), we have;

KE = 0.5 x m x v²

KE = 0.5*93000*1000*(16.448)²

= 12.58 x $10^{9}$ J

= 12.58 GJ

6 0

3 years ago

A 0.30 kg softball has a velocity of 15 m/s at an angle of 35 degrees below the horizontal just before making contact with the b

jarptica [38.1K]

Answer:

a) 5.03 kg m/s

b) 10.03 kg m/s

Explanation:

Hi!

Let us consider the origin of coordinates at the pitcher, and pointing directly towards the initial direction of the ball. Therefore, the angle of the velocity with respect to the x axis is -35° (below the horizontal).

The components of the initial momentum are:

px = (0.3 kg)(15 m/s) cos(-35° ) = 4.5 cos(-35° ) kg m/s = 3.69 kg m/s

py = 4.5 sin(-35° ) kg m/s = -2.581 kg m/s

The final momentum will be:

a)

pfx = 0

pfy = - (20 m/s) (0.3 kg) = -6 m/s

And the difference in momentum is:

dpy = pfy - py = -3.419 kg m/s

dpx = pfx - px = -3.69 kg m/s

And its magnitude:

dp = √(dpy^2 + dpx^2) = 5.03 kg m/s

b)

pfx = -6 kg m/s

pfy = 0

And the difference in momentum is:

dpx = pfx - px = -9.69 kg m/s

dpy = pfy - py = 2.581 kg m/s

And its magnitude:

dp = √(dpy^2 + dpx^2) = 10.03 kg m/s

6 0

3 years ago

Machines can be made more efficient by reducing ___________________.

Aleks04 [339]

Answer:

friction

wubba lubba dub dub

3 0

3 years ago

What is the final velocity, in meters per second, of a freight train that accelerates at a rate of 0.085 m/s2 for 7.5 min, start

Llana [10]

Answer:

$v_f=41.65\frac{m}{s}$

Explanation:

The final velocity is given by the following kinematic equation:

$v_f=v_0+at$

Here, $v_0$ is the initial velocity, a is the body's acceleration and t is the motion time. We have to convert the time to seconds:

$7.5min*\frac{60s}{1min}=450s$

Now, we calculate the final velocity:

$v_f=3.4\frac{m}{s}+(0.085\frac{m}{s^2}(450s))\\v_f=41.65\frac{m}{s}$

4 0

3 years ago