A decrease in the magnitude of velocity is called

At a location, rainwater runs off the face of a cliff that is 15 meters high. How is the face of the cliff likely to change over

stiv31 [10]

Weathering by water would cut its top deeper than the middle

5 0

3 years ago

A tennis player serves a tennis ball such that it is moving horizontally when it leaves the racquet. When the ball travels a hor

finlep [7]

Answer:

The initial velocity is 38.46 m/s.

Explanation:

The horizontal distance travel by the tennis ball = 13 m

The height at which the tennis ball dropped = 56 cm

Now calculate the initial speed of tennis ball.

The vertical velocity is zero.

Below is the calculation. Here, first convert centimetre into kilometre. So, height at which ball dropped is 0.56 km.

$v = \sqrt{2 \times 9.8 \times 0.56} = 3.32 m/s \\$

$t = \frac{3.32}{9.8} = 0.338s \\$

$Ux \times t = 13 \\$

$Ux = \frac{13}{0.338} = 38.46 m/s = Initial velocity.$

8 0

3 years ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin

lyudmila [28]

Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
$W= \int\limits^{0.540m}_{0} {F} \, dx = \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =$
$=16000x+10000 \frac{x^2}{2} - 26000 \frac{x^3}{3}$
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
$W=16000(0.540m)+10000 \frac{(0.540m)^2}{2} - 26000 \frac{(0.540m)^3}{3} =$
$=8733 J=8.73 kJ$

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
$W=16000(0.95m)+10000 \frac{(0.95m)^2}{2} - 26000 \frac{(0.95m)^3}{3} =$
$=12280 J=12.28 kJ$

5 0

3 years ago

A large, cylindrical water tank with diameter 3.60 m is on a platform 2.00 m above the ground. The vertical tank is open to the

zysi [14]

To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.

The volume of a tank is given by

$V = \frac{\pi d^2}{4}h$

Where

d = Diameter

h = Height

Considering that there are two stages, let's define the initial and final volume as,

$V_0 = \frac{\pi d^2}{4}H$

$V_f = \frac{\pi d^2}{4}h$

We know as well by definition that

$1gal = 3.785*10^{-3}m^3$

Then we have for the statement that

$V_f = V_0 -1gal$

$V_f = V_0 - 3.785*10^{-3}$

Replacing the previous data

$\frac{\pi d^2}{4}h = \frac{\pi d^2}{4}H- 3.785*10^{-3}$

$\frac{\pi (3.6)^2}{4}h = \frac{\pi (3.6)^2}{4}(2)- 3.785*10^{-3}$

Solving to get h,

$h = 1.99963m$

Therefore the change is

$\Delta h = H-h$

$\Delta h = 2- 1.99963$

$\Delta h = 3.7*10^{-4}m=0.37mm$

Therefore te change in the height of the water in the tank is 0.37mm

4 0

3 years ago

(b) Two electrons in the same atom both have n = 3 and I = 1. Assume the

wolverine [178]

Answer:

We are social beings and we are going to be able to make a paper gun very very small it want to be a hacker in real life and we are going to be able to make

8 0

3 years ago