Who is the silentest that revised the atomic modle

A ball is projected horizontally from the top of a bertical building 25.0m above the ground level with an initial velocity of 8.

kirill115 [55]

Answer:

Solution given:

height [H]=25m

initial velocity [u]=8.25m/s

g=9.8m/s

now;

a. How long is the ball in flight before striking the ground?

Time of flight =?

Now

Time of flight= $\sqrt{\frac{2H}{g}}$

substituting value

= $\sqrt{\frac{2*25}{9.8}}$
=2.26seconds

<h3>the ball is in flight before striking the ground for 2.26seconds.</h3>

b. How far from the building does the ball strike the ground?

Horizontal range=?

we have

Horizontal range=u* $\sqrt{\frac{2H}{g}}$

=8.25*2.26
=18.63m

<h3>The ball strikes 18.63m far from building. </h3>

7 0

2 years ago

If you were making an electric device and needed a piece that would easily transmit an electric charge and also not react with o

loris [4]

Answer:

This question is incomplete

Explanation:

This question is incomplete because of the absence of options. However, one material that is good candidate for conducting electricity without reacting with other materials is metallic vanadium dioxide. This is because of the inability of this electrical conductor to conduct heat (an unusual property for all other electrical conductors) and thus makes it difficult for it to react with other materials (since an increase in temperature increases possibility of a reaction).

6 0

3 years ago

A bowling ball with a momentum of 18kg-m/s strikes a stationary bowling pin. After the collision, the ball has a momentum of 13k

Veronika [31]

Answer:

$14.98\ \text{kg m/s}$

$45.26^{\circ}$

Explanation:

$P_1$ = Initial momentum of the pin = 13 kg m/s

$P_i$ = Initial momentum of the ball = 18 kg m/s

$P_2$ = Momentum of the ball after hit

$55^{\circ}$ = Angle ball makes with the horizontal after hitting the pin

$\theta$ = Angle the pin makes with the horizotal after getting hit by the ball

Momentum in the x direction

$P_i=P_1\cos55^{\circ}+P_2\cos\theta\\\Rightarrow P_2\cos\theta=P_i-P_1\cos55^{\circ}\\\Rightarrow P_2\cos\theta=18-13\cos55^{\circ}\\\Rightarrow P_2\cos\theta=10.54\ \text{kg m/s}$

Momentum in the y direction

$P_1\sin55=P_2\sin\theta\\\Rightarrow P_2\sin\theta=13\sin55^{\circ}\\\Rightarrow P_2\sin\theta=10.64\ \text{kg m/s}$

$(P_2\cos\theta)^2+(P_2\sin\theta)^2=P_2^2\\\Rightarrow P_2=\sqrt{10.54^2+10.64^2}\\\Rightarrow P_2=14.98\ \text{kg m/s}$

The pin's resultant velocity is $14.98\ \text{kg m/s}$

$P_2\sin\theta=10.64\\\Rightarrow \theta=sin^{-1}\dfrac{10.64}{14.98}\\\Rightarrow \theta=45.26^{\circ}$

The pin's resultant direction is $45.26^{\circ}$ below the horizontal or to the right.

4 0

2 years ago

Trên bóng đèn có ghi là 12v -6w. Khi đèn sáng bình thường thì dòng điện chạy qua đèn có cường độ là

Annette [7]

Answer:

Xin lỗi, ở đây không có ai nói tiếng Việt, nhưng bạn có thể cuộn hết cỡ xuống dưới để tìm một trang web cho não biết nói tiếng Việt

Explanation:

4 0

2 years ago

Calculate the maximum capillary rise/fall of mercury in a 0.5 mm radius glass capillary. Assume that the surface tension for mer

tekilochka [14]

Answer: 0.01 m

Explanation: The formulae for capillarity rise or fall is given below as

h = (2T×cosθ)/rpg

Where θ = angle mercury made with glass = 50°

T = surface tension = 0.51 N/m

g = acceleration due gravity = 9.8 m/s²

r = radius of tube = 0.5mm = 0.0005m

p = density of mercury.

h = height of rise or fall

From the question, specific gravity of density = 13.3

Where specific gravity = density of mercury/ density of water, where density of water = 1000 kg/m³

Hence density of mercury = 13.3×1000 = 13,300 kg/m³.

By substituting parameters, we have that

h = 2×0.51×cos 50/0.0005×9.8×13,300

h = 0.6556/65.17

h = 0.01 m

8 0

2 years ago