All categories

3 years ago

If a ball is launched horizontally at 40 m/s

Physics

1 answer:

vaieri [72.5K]3 years ago

4 0

Answer:

40m/s

Explanation:

The horizontal component of velocity remains constant because there are no external forces in that direction

By applying motion equations, V= U+ at

where ,

v - final velocity
u - initial velocity
a-acceleration
t - time

v = u +at

As no force act on the ball ( we neglect air resistance here) no acceleration is seen,

So v = u = 40m/s

You might be interested in

A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

4 0

2 years ago

Calculate the de Broglie wavelength of (a) a mass of 1.0 g traveling at 1.0 m s−1 , (b) the same, traveling at 1.00 × 105 km s−1

lesantik [10]

Answer:

a) $\lambda=6.63\times10^{-31}m$

b) $\lambda=6.63\times10^{-39}m$

c) $\lambda=9.97\times10^{-11}m$

d) $\lambda=4.03\times10^{-36}$ m

e)λ=∞

Explanation:

De Broglie discovered that an electron or other mass particles can have a wavelength associated, and that wavelength (λ) is:

$\lambda=\frac{h}{P}=\frac{h}{mv}$

with h the Plank's constant ( $6.63\times10^{-34}\frac{m^{2}kg}{s}$ ) and P the momentum of the object that is mass (m) times velocity (v).

a) $\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}kg*1.0)}$

$\lambda=6.63\times10^{-31}m$

b) $\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}*(1.00\times10^{8}))}$

$\lambda=6.63\times10^{-39}m$

c) $\lambda=\frac{6.63\times10^{-34}}{(6.65\times10^{-27}*1000)}$

$\lambda=9.97\times10^{-11}m$

d) $\lambda=\frac{6.63\times10^{-34}}{(74*2.22)}$

$\lambda=4.03\times10^{-36}$ m

e) $\lambda=\frac{6.63\times10^{-34}}{(74*0)}$

λ=∞

6 0

3 years ago

Changes of state

Mrrafil [7]

Answer:

Option A

Explanation:

At segment T-U, the substance changes from a liquid to a gas and does not change temperature.

The reason is because latent heat of vaporisation allows for the absorption of heat in the change of state and temperature remains constant until it has fully changed state.

4 0

1 year ago

Read 2 more answers

How do you find input energy with only the efficiency and output energy given?

iris [78.8K]

Answer:

power output/transformer efficiency multiply by 100%

Explanation:

derrive the formula from efficiency=power output/power input multiply by 100%

8 0

3 years ago

Given a second class lever with a distance of 5.00 feet from the fulcrum to the effort and a distance of 33.0 inches from the re

leva [86]

Answer:

The correct answer is C. 45.5 lbs.

Explanation:

In a second class lever, the load is located between the point in which the force is exerted and the fulcrum.

The formula for any problem involving a lever is:

$F_ed_e=F_ld_l$

Where F_e is the effort force, d_e is the total length of the lever, F_l is the load that can be lifted and d_l is the distance between the point of the effort and the fulcrum.

The parameter of the formula that you need is F_l:

$F_l=\frac{F_ed_e}{d_l}$

The conversion from feet to inches is 1 ft is equal to 12 inches. In this case, 5 ft are equal to 60 inches.

$F_l=\frac{25*60}{33}$

F_l=45.5 lbs

7 0

3 years ago