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d1i1m1o1n [39]
2 years ago
10

A bullet is fired at an angle θ above the horizontal with an initial velocity of 800 m/s from the top of an 80 m high tower. Wha

t value of θ will give the maximum horizontal range?
Physics
1 answer:
Irina-Kira [14]2 years ago
6 0

Answer:

pahingi po ng pic pls para masagutang kopo iyan

You might be interested in
510 g squirrel with a surface area of 935 cm2 falls from a 4.8-m tree to the ground. Estimate its terminal velocity. (Use the dr
Yuki888 [10]

Answer:

The terminal velocity is v_t  =17.5 \ m/s

Explanation:

From the question we are told that

       The mass of the squirrel is  m_s  =  50\ g  =  \frac{50}{1000} =  0.05 \  kg

      The surface area is   A_s =  935 cm^2  =  \frac{935}{10000} = 0.0935 \ m^2

       The height of fall is  h =4.8 m

        The length of the prism is l =  23.2 = 0.232 \ m

          The width of the prism is w =  11.6 =  0.116 \ m

 

The terminal velocity is mathematically represented as

       v_t  =  \sqrt{\frac{2 * m_s *  g }{\dho_s * C  * A } }

Where \rho  is the density of a rectangular prism with a constant values of \rho  =  1.21 \ kg/m^3

            C is the drag coefficient for a horizontal skydiver with a value = 1

            A  is the area of the prism the squirrel is assumed to be which is mathematically represented as

      A =  0.116 * 0.232

       A =  0.026912 \ m^2

 substituting values

      v_t  =  \sqrt{\frac{2 * 0.510 *  9.8 }{1.21 * 1  * 0.026912 } }

     v_t  =17.5 \ m/s

       

7 0
3 years ago
If the person drops box from 3.8 m how much energy is transferred from potential energy to kinetic energy
kotykmax [81]

Answer:

Kinetic energy

When work is done the energy is transferred from one type to another. This transferred energy may appear as kinetic energy.

For example, when you pedal your bicycle so that its speed increases, you are doing work to transfer chemical energy from your muscles to the kinetic energy of the bicycle.

Kinetic energy is the energy an object possesses by virtue of its movement. The amount of kinetic energy possessed by a moving object depends on the mass of the object and its speed. The greater the mass and the speed of the object the greater its kinetic energy.

The kinetic energy Ek of an object of mass m at a speed v is given by the relationship

{E_k} = \frac{1}{2}m{v^2}

m is the mass of the object in kilograms ( kg) and v is the speed of the object in metres per second ( m\,s^{-1}).

Explanation:

When work is done on an object it may also lead to energy being transferred to the object in the form of gravitational potential energy of the object.

Gravitational potential energy is the energy an object has by virtue of its position above the surface of the Earth. When an object is lifted, work is done. When work is done in raising the height of an object, energy is transferred as a gain in the gravitational potential energy of the object.

For example, suppose you lift a suitcase of mass m through a height h. The weight W of the suit case is a downward force of size mg. In lifting the suitcase, you would have to pull upwards on it with a force equal in size to its weight, mg.

Two suitcases. One has a green force arrow pointing up labelled F and a purple force arrow pointing down labelled 'Weight = mg'. The other case is raised by a height labelled h.

Suitcases with forces and height labelled

When this force (equal to the weight mg, but upwards) is applied to the suitcase over the distance h:

Work\,done=force\,\times\,distance\,upwards=mg\,\times\,h

This energy is transferred to potential energy when raising the object through a known height.

Energy = mass \times gravitational\,field\,strength \times height

E = m \times g \times h

This is the relationship used to calculate gravitational potential energy.

{E_p} = mgh

where m is the mass of the object in kilograms (kg), g is the gravitational field strength, (for positions near the surface of the Earth g = 9∙8 newtons per kilogram ( N kg ^{-1} and h is the height above the surface of the Earth in metres ( m).

8 0
3 years ago
A 87.0 kg astronaut is working on the engines of a spaceship that is drifting through space with a constant velocity. The astron
Ket [755]

Answer:

259.62521 seconds

Explanation:

m_1 = Mass of astronaut = 87 kg

m_2 = Mass of wrench = 0.57 kg

v_1 = Velocity of astronaut

v_2 = Velocity of wrench = 22.4 m/s

Here, the linear momentum is conserved

m_1v_1=m_2v_2\\\Rightarrow v_1=\frac{m_2v_2}{m_1}\\\Rightarrow v_1=\frac{0.57\times 22.4}{87}\\\Rightarrow v_1=0.14675\ m/s

Time = Distance / Speed

Time=\frac{38.1}{0.14675}=259.62521\ s

The time taken to reach the ship is 259.62521 seconds

4 0
3 years ago
Please help me
g100num [7]

Answer:

0.3956

Explanation:

Newton's 2nd law of motion says that Force = Mass*Acceleration (f=ma) so to find the force used on the football you multiply it's mass by its acceleration.

0.43*0.92 = 0.3956.

0.4 if you round

5 0
3 years ago
A train is moving in a straight railway where it covered one third of the distance with
Alexxandr [17]

Answer:

<em>The average speed of the train is 45 km/h</em>

Explanation:

<u>Speed</u>

It's defined as the distance (d) per unit of time (t) traveled by an object. The formula is:

\displaystyle v=\frac{d}{t}

Let's call x the total distance covered by the train. It covered d1=1/3x with a speed of v1=25 km/h. The time taken is calculated solving for t:

\displaystyle t_1=\frac{d_1}{v_1}

\displaystyle t_1=\frac{1/3x}{25}

\displaystyle t_1=\frac{x}{75}

Now the rest of the distance:

d2 = x - 1/3x = 2/3x

Was covered at v2=75 km/h. Thus the time taken is:

\displaystyle t_2=\frac{d_2}{v_2}

\displaystyle t_2=\frac{2/3x}{75}

\displaystyle t_2=\frac{2x}{225}

The total time is:

\displaystyle t_t=\frac{x}{75}+\frac{2x}{225}

\displaystyle t_t=\frac{3x}{225}+\frac{2x}{225}

\displaystyle t_t=\frac{5x}{225}

Simplifying:

\displaystyle t_t=\frac{x}{45}

The average speed is the total distance divided by the total time:

\displaystyle \bar v=\frac{x}{\frac{x}{45}}

Simplifying:

\boxed{\displaystyle \bar v=45\ km/h}

The average speed of the train is 45 km/h

5 0
3 years ago
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