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mr Goodwill [35]
2 years ago
11

You are in a boat that is 1400 kg. The current is 1500N and is pushing you back. If you have an acceleration of 3 m/s^2 what is

the force coming out of the boat's engines?
Physics
1 answer:
aleksandrvk [35]2 years ago
5 0

Answer: The force of the engines is 5700N

Explanation:

By the second Newton's law, we have that:

F = m*a

Force equals mass times acceleration.

In this case, we have:

m = 1400kg

a = 3 m/s^2

And the force will be equal to the force of the engines, f, plus the force of  the current, -1500 N (because this is pushing you back, so it is in the opposite direction than f), then we have:

F = f - 1500N

Then we have the equation:

f - 1500N = 1400kg*3m/s^2 = 4200N

f - 1500N = 4200N

f = 4200N + 1500N = 5700N

The force of the engines is 5700N

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Vlada [557]

Answer:

Vmax=11.53 m/s

Explanation:

from conservation of energy

      E_A} =E_{B}

     Spring potential energy =potential energy due to elevation

   0.5*k*x²= mg(h_{B}-h_{A} )=mgh

   0.5*k*2.3²= 430*9.81*6

         k=9568.92 N/m

For safety reason

                                 k"=1.13 *k= 1.13*9568.92

                                    k"=10812.88 N/m

agsin from conservation of energy

      E_A} =E_{C}

    spring potential energy=change in kinetic energy

   0.5*k"*x²=0.5*m*V_{max}^{2}

      10812.88 *2.3²=430*V_{max}^{2}

           V_{max}=11.53 m/s

5 0
3 years ago
A block of size 20m x 10 mx 5 m exerts a force of 30N. Calculate the
Orlov [11]

Answer:

We know that force applied per unit area is called pressure.

Pressure = Force/ Area

When force is constant than pressure is inversely proportional to area.

1- Calculating the area of three face:

A1 = 20m x 10 m =200 Square meter

A2 = 10 mx 5 m = 50 Square meter

A3 = 20m x 5 m = 100 Square meter

Therefore A1 is maximum and A2 is minimum.

2- Calculate pressure:

P = F/ A1 = 30 / 200 = 0.15 Nm⁻²  ( minimum pressure)

P = F / A2 = 30 / 50 = 0.6 Nm⁻²   ( maximum pressure)

Hence greater the area less will be the pressure and vice versa.

3 0
2 years ago
A vertical spring with a spring constant of 420 N/m is mounted on the floor. From directly above the spring, which is unstrained
mixas84 [53]

Answer:

19.53 cm

Explanation:

The computation of the height is as follows:

Here we applied the conservation of the energy formula

As we know that

P.E of the block = P.E of the spring

 m g h = ( 1 ÷ 2) k x^2

where

m = 0.15

g = 9.81

k = 420

x = 0.037

So now put the values to the above formula

(0.15) (9.81) (h) = 1 ÷2 × 420 × (0.037)^2

1.4715 (h) = 0.28749

h = 0.19537 m

= 19.53 cm

3 0
2 years ago
We wrap a light, nonstretching cable around a 8.00 kg solid cylinder with diameter of 30.0 cm. The cylinder rotates with negligi
antoniya [11.8K]

Answer:

h = 16.67m

Explanation:

If the kinetic energy of the cylinder is 510J:

Kc=510=1/2*Ic*\omega c^2

\omega c=\sqrt{510*2/Ic}

Where the inertia is given by:

Ic=1/2*m_c*R_c^2=1/2*(8)*(0.15)^2=0.0225kg.m^2

Replacing this value:

\omega c=106.46rad/s

Speed of the block will therefore be:

V_b=\omega_c*R_c=106.46*0.15=15.969m/s

By conservation of energy:

Eo = Ef

Eo = 0

Ef = 510+1/2*m_b*V_b^2-m_b*g*h

So,

0 = 510+1/2*m_b*V_b^2-m_b*g*h

Solving for h we get:

h=16.67m

3 0
3 years ago
Which of the following is NOT an example of a correct interaction between the
makkiz [27]

Answer:

WE NEED ANSWER CHOICES

Explanation:

3 0
3 years ago
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