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2 years ago

an eagle carries a fish up 50 m into the sky using 90 n of force. how much work did the eagle do on the fish

Physics

2 answers:

Colt1911 [192]2 years ago

6 0

Answer:

The answer is 4,500

Explanation:

W = FD

W = 90 x 50

W = 4,500

Mashutka [201]2 years ago

4 0

The answer would be 1.8 acceleration

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A car travels at 2000 meters east in 40 seconds. What is the cars velocity

DerKrebs [107]

Answer:

50 ms⁻¹

Explanation:

Given that,

Distance Travelled =2000 meters

Time Taken =40 seconds

velocity = ?

we know,

velocity = Distance / Time

=(2000 / 40) ms⁻¹

=50 ms⁻¹

5 0

3 years ago

Two drums of the same size and same height are taken.

Gelneren [198K]

Answer:

i) The pressure acting on the base of B will be half the pressure acting on the base of A

ii) The pressure acting on the base of B will be the same as the pressure acting on the base of A

iii) The pressure on the base of drum A will be slightly less than the pressure on the base of drum B

Explanation:

The pressure acting on the base of the drum, P = h·ρ·g

Where;

h = The level of the liquid in the drum

$h_{max}$ = The height of the drums

ρ = The density of the liquid in the drum

g = The acceleration due to gravity ≈ 9.81 m/s²

i) If A is completely filled, we have $h_A$ = $h_{max}$

Therefore, $P_A$ = $h_{max}$ × $\rho_{liquid}$ ×g

If B is half filled, we have, $h_B$ = (1/2)· $h_{max}$

$P_B$ = (1/2) × $h_{max}$ × $\rho_{liquid}$ ×g

Therefore, $P_B$ = (1/2) × $P_A$

The pressure acting on the base of B will be half the pressure acting on the base of A

ii) If both A and B are each filled with water (the same liquid), then the pressure on their bases will be $P_A$ = $h_{max}$ × $\rho_{water}$ ×g = $P_B$ , the same, given that the acceleration due to gravity, g, is constant and the same in Nepal and India

iii) If A is filled with water, and B is filled with salty water, we have that, the density of salty water is slightly higher than water, therefore, we get;

$P_A$ = $h_{max}$ × $\rho_{water}$ ×g < $P_B$ =

The pressure on the base of drum A will be less than the pressure on the base of drum B.

3 0

3 years ago

How long will it take to go 150 km [E] traveling at 50 km/hr?

pantera1 [17]

It will take 3 hours

7 0

3 years ago

Read 2 more answers

A girl of mass m1=60 kilograms springs from a trampoline with an initial upward velocity of v1=8.0 meters per second. At height

AleksandrR [38]

a) 5.0 m/s

This first part of the problem can be solved by using the conservation of energy. In fact, the mechanical energy of the girl just after she jumps is equal to her kinetic energy:

$E_i=\frac{1}{2}m_1v_1^2$

where m1 = 60 kg is the girl's mass and v1 = 8.0 m/s is her initial velocity.

When she reaches the height of h = 2.0 m, her mechanical energy is sum of kinetic energy and potential energy:

$E_f = \frac{1}{2}m_1 v_2 ^2 + m_1 gh$

where v2 is the new speed of the girl (before grabbing the box), and h = 2.0m. Equalizing the two equations (because the mechanical energy is conserved), we find

$\frac{1}{2}m_1 v_1^2 = \frac{1}{2}m_1 v_2 ^2 + m_1 gh\\v_1^2 = v_2^2 +2gh\\v_2 = \sqrt{v_1^2 -2gh}=\sqrt{(8.0 m/s)^2-(2)(9.8 m/s^2)(2.0 m)}=5.0 m/s$

b) 4.0 m/s

After the girl grab the box, the total momentum of the system must be conserved. This means that the initial momentum of the girl must be equal to the total momentum of the girl+box after the girl catches the box:

$p_i = p_f\\m_1 v_2 = (m_1 + m_2) v_3$

where m2 = 15 kg is the mass of the box. Solving the equation for v3, the combined velocity of the girl+box, we find

$v_3 = \frac{m_1 v_2}{m_1 + m_2}=\frac{(60 kg)(5.0 m/s)}{60 kg+15 kg}=4 m/s$

c) 2.8 m

We can use again the law of conservation of energy. The total mechanical energy of the girl after she catches the box is sum of kinetic energy and potential energy:

$E_i = \frac{1}{2}(m_1+m_2) v_3^2 + (m_1+m_2)gh=\frac{1}{2}(75 kg)(4 m/s)^2+(75 kg)(9.8 m/s^2)(2.0m)=2070 J$