Ask question

Ask question

All categories

3 years ago

5

Tom Cruise jumped from one building to other building while filming the roof chase scene in Mission: Impossible - Fallout. He di

d not land on the roof of the other building safely and broke his ankle. If we assume the heights of the two building are the same and the distance between the buildings is 5 m, what is the minimum speed to land on the roof of the other building? Assume the jumping angle is 15 degrees and the air friction is negligible. (15 points) Vo =? O = 15° D = 5 m

1 answer:

OLga [1]3 years ago

3 0

Answer:

$v_0=9.9\ m.s^{-1}$

Explanation:

Given:

angle of launch of projectile from horizontal, $\theta=15^{\circ}$

range of projectile, $R=5\ m$

<u>We have formula for the range of projectile:</u>

$R=\frac{v_0^2\times sin\ 2\theta}{g}$

putting the respective values

$5=\frac{v_0^2\times sin\ 30^{\circ}}{9.8}$

$v_0=9.9\ m.s^{-1}$ is the velocity with which Tom should jump to land on the other roof.

You might be interested in

A graph titled Distance as a Function of Time with horizontal axis time (seconds) and vertical axis distance (meters). A straigh

yanalaym [24]

Answer:

3 m

Explanation:

6 0

2 years ago

Read 2 more answers

An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standar

defon

Answer:

Explanation:

Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.

To objective is to find the:

(i) required heat exchanger area.

(ii) flow rate to be maintained in the evaporator.

Given that:

water temperature = 300 K

At a reasonable depth, the water is cold and its temperature = 280 K

The power output W = 2 MW

Efficiency $\zeta$ = 3%

where;

$\zeta = \dfrac{W_{out}}{Q_{supplied }}$

$Q_{supplied } = \dfrac{2}{0.03} \ MW$

$Q_{supplied } = 66.66 \ MW$

However, from the evaporator, the heat transfer Q can be determined by using the formula:

Q = UA(L MTD)

where;

$LMTD = \dfrac{\Delta T_1 - \Delta T_2}{In (\dfrac{\Delta T_1}{\Delta T_2} )}$

Also;

$\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K$

$\Delta T_2 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_2 = 292 -290 \\ \\ \Delta T_2 = 2\ K$

$LMTD = \dfrac{10 -2}{In (\dfrac{10}{2} )}$

$LMTD = \dfrac{8}{In (5)}$

LMTD = 4.97

Thus, the required heat exchanger area A is calculated by using the formula:

$Q_H = UA (LMTD)$

where;

U = overall heat coefficient given as 1200 W/m².K

$66.667 \times 10^6 = 1200 \times A \times 4.97 \\ \\ A= \dfrac{66.667 \times 10^6}{1200 \times 4.97} \\ \\ \mathbf{A = 11178.236 \ m^2}$

The mass flow rate:

$Q_{H} = mC_p(T_{in} -T_{out} ) \\ \\ 66.667 \times 10^6= m \times 4.18 (300 -292) \\ \\ m = \dfrac{ 66.667 \times 10^6}{4.18 \times 8} \\ \\ \mathbf{m = 1993630.383 \ kg/s}$

3 0

3 years ago

A car driving at 20 m/s accelerates continuously 2m/s2 for 3 seconds. What is its final velocity

Lesechka [4]

Answer: V= u+ at

V= final velocity

u=initial velocity

a=acceleration

t=time taken

V= 20 + 2*3

V= 26m/s

Explanation:

5 0

3 years ago

In a chemical equation, the chemicals that react are considered . In a chemical equation, the chemicals that are produced are co

vovangra [49]

Answer

Hi,

In a chemical equation, chemicals that react are the reactants, while chemicals that are produced are the products/by products. Both sides of the equation must be balanced.

Explanation

When writing a chemical equation, reactants reacts to produce products. For example in the equation for formation of water, hydrogen combines with oxygen as 2H₂ +O₂→2H₂O where the first part before the arrow represent the reactants and the next part after the arrow are the products. Reactants are on the left where as products are on the right.Coefficient 2, in this cases is used for balancing the equation.

Good luck!

4 0

3 years ago

Read 2 more answers

An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

4 0

3 years ago