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serious [3.7K]
3 years ago
15

Trained dolphins are capable of a vertical leap of 7.0m straight up from the surface of the water-an impressive feat.Suppose you

could train a dolphin to launch itself out of the water at this same speed but at an angle.What maximum horizontal range could the dolphin achieve?
Physics
1 answer:
anzhelika [568]3 years ago
5 0

Answer:

   θ = 45º

Explanation:

To find the solution, let's use the projectile launch equation

    R = vo² sin 2θ / g

Where vo is the initial speed of the dolphin, T is the angle of the jump and g the gravity acceleration.

To obtain a maximum range, the sine T function must be 1,

     sin 2θ = 1

     2θ = sin⁻¹ 1

     2θ = 90

     θ = 45º

Therefore, the dolphin should jump at an angle of 45º from the horizontal

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Light with an intensity of 1 kW/m2 falls normally on a surface and is completely absorbed. The radiation pressure is
kobusy [5.1K]

Answer:

The radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

Explanation:

Given;

intensity of light, I = 1 kW/m²

The radiation pressure of light is given as;

Radiation \ Pressure = \frac{Flux \ density}{Speed \ of \ light}

I kW = 1000 J/s

The energy flux density = 1000 J/m².s

The speed of light = 3 x 10⁸ m/s

Thus, the radiation pressure of the light is calculated as;

Radiation \ pressure = \frac{1000}{3*10^{8}} \\\\Radiation \ pressure =3.33*10^{-6} \ Pa

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6 0
3 years ago
A person walks 9.0 km directly east and then turns left and heads directly north for 12.0 km. What is his displacement from the
Readme [11.4K]

Explanation:

Given that,

A person walks 9.0 km directly east and then turns left and heads directly north for 12.0 km.

We need to find his displacement from the starting position.

We know that,

Displacement = shortest path covered

D=\sqrt{9^2+12^2} \\\\D=15\ km

For direction,

\theta=\tan^{-1}(\dfrac{d_y}{d_x})\\\\\theta=\tan^{-1}(\dfrac{12}{9})\\\\= $$53.13^{\circ}

Hence, this is the required solution.

5 0
2 years ago
what was the initial temperature is 250 calories reply .1 kg of gold the final temperature of the gold was 175°c ? the specific
weqwewe [10]

Answer:

115.2^{\circ}C

Explanation:

When an amount of energy Q is supplied to a substance of mass m, the temperature of the substance increases by \Delta T, according to the equation

Q=mC_s \Delta T

where C_s is the specific heat capacity of the substance.

In this problem, we have:

Q=250 \cdot 4.184 =1046 J is the amount of heat supplied to the sample of gold

m = 0.1 kg = 100 g is the mass of the sample

C_s = 0.175 J/gC is the specific heat capacity of gold

Solving for \Delta T, we find the change in temperature

\Delta T = \frac{Q}{m C_s}=\frac{1046}{(100)(0.175)}=59.8^{\circ}

And since the final temperature was

T_f = 175^{\circ}

The initial temperature was

T_i = T_f - \Delta T= 175 -59.8=115.2^{\circ}C

3 0
3 years ago
URGENT What does Newton's third law say about why momentum is conserved in collisions?
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3 years ago
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What force always attracts objects to each other ?
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I'm pretty your it's magnetism though. Like magnets 
4 0
2 years ago
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