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densk [106]
3 years ago
8

A 1000 kg car is rolling slowly across a level surface at 1m/s, heading toward a group of small innocent children.

Physics
2 answers:
Ne4ueva [31]3 years ago
7 0

Answer:

The force is -1620.73 N.

Explanation:

Given that,

Mass of car = 1000 kg

Velocity = 1 m/s

Distance = 2 m

Angle = 30°

We need to calculate the force

Using formula of work done

W_{net}=K_{f}-K_{i}

F\times d=K_{f}-K_{i}

Fd\cos\theta=-K_{i}

F=\dfrac{-K_{f}}{d\cos\theta}

F=-\dfrac{\dfrac{1}{2}mv^2}{d\cos\theta}

Put the value into the formula

F=-\dfrac{\dfrac{1}{2}\times1000\times(1)^2}{2\times\cos30}

F=-1620.73\ N

Hence, The force is -1620.73 N.

AnnyKZ [126]3 years ago
4 0

Answer:

Force, F = 288.67 N

Explanation:

Mass of the car, m = 1000 kg

Initial speed of the car, u = 1 m/s

The car and push on the hood at an angle of 30° below horizontal, \theta=30^{\circ}

Distance, d = 2 m

Let F is the force must you push to stop the car. using work energy theorem to find it. According to this theorem, the work done is equal to the change in kinetic energy as :

W=\dfrac{1}{2}m(v^2-u^2)

F\times d=\dfrac{1}{2}m(v^2-u^2)

It stops, v = 0

Fd\ cos\theta=\dfrac{1}{2}m(u^2)      

F=\dfrac{\dfrac{1}{2}m(u^2)}{d\ cos\theta}

F=\dfrac{\dfrac{1}{2}\times 1000\times (1)^2}{2\ cos(30)}

F = -288.67 N

So, the force required to push to stop the car is 288.67 N. Hence, this is the required solution.                                            

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The mass of moon is 1/100 times and its radius 1/4 times that of earth as a result the gravitational attraction on the moon is about one sixth when compared to earth.

7 0
2 years ago
What should Ben wear to best protect himself during the experiment?
GarryVolchara [31]
Goggles, a lab coat, chemical gloves, and close- toed shoes
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3 years ago
Read the scenario and solve these two problems.
Burka [1]

Answers:

a) 5400000 J

b) 45.92 m

Explanation:

a) The kinetic energy K of an object is given by:

K=\frac{1}{2}mV^{2}

Where:

m=12000 kg is the mass of the train

V=30 m/s is the speed of the train

Solving the equation:

K=\frac{1}{2}(12000 kg)(30 m/s)^{2}

K=5400000 J This is the train's kinetic energy at its top speed

b) Now, according to the Conservation of Energy Law, the total initial energy is equal to the total final energy:

E_{i}=E_{f}

K_{i}+P_{i}=K_{f}+P_{f}

Where:

K_{i}=5400000 J is the train's initial kinetic energy

P_{i}=0 J is the train's initial potential energy

K_{f}=0 J is the train's final kinetic energy

P_{f}=mgh is the train's final potential energy, where g=9.8 m/s^{2} is the acceleration due gravity and h is the height.

Rewriting the equation with the given values:

5400000 J=(12000 kg)(9.8 m/s^{2})h

Finding h:

h=45.918 m \approx 45.92 m

7 0
3 years ago
Read 2 more answers
A 3-m-high, 7-m-wide rectangular gate is hinged at the top edge and is restrained by a fixed ridge. Determine the hydrostatic fo
Shalnov [3]

Answer:

The Hydrostatic force is   F  =  137.2 kN

The location of pressure center is  Z  = 1.333 \ m  

Explanation:

From the question we are told that

   The height of the gate is  h =  3 \ m

     The weight of the gate is  w =  7 \  m

      The height of the water is  h_w  =  2 \ m

       The density of water is \rho_w  =  1000 \ kg/m^3

Note used h_w for height of water and height of gate immersed by water since both have the same value

The area of the gate immersed in water  is mathematically represented as

         A =  h_w  * w

substituting values

         A =  2*  7

         A =  14  \ m^2

The hydrostatic force is mathematically represented as

          F  =  \rho_w * g * h_f * A

Where

            h_f =h-  h_w

           h_f =3 -2

           h_f = 1\ m  

So  

              F  =  1000 * 9.8 * 1 * 14

            F  =  137.2 kN

The center of pressure is mathematically represented as

        Z  =  h_f + \frac{I_g}{h_f * A}

Where I_g is the moment of inertia of the gate which mathematically represented as

            I_g =  \frac{w * h_w^2}{12}

The h_w is the height of gate immersed in water

            I_g =  \frac{7  * 2^2 }{12}

             I_g = 4.667\ kg  m^2

Thus  

        Z  = 1  + \frac{4.66}{1 * 14}

        Z  = 1.333 \ m

3 0
3 years ago
f a single circular loop of wire carries a current of 45 A and produces a magnetic field at its center with a magnitude of 1.50
Lelu [443]

Answer:

Radius of the loop is 0.18 m or 18 cm

Explanation:

Given :

Current flowing through the wire, I = 45 A

Magnetic field at the center of the wire, B = 1.50 x 10⁻⁴ T

Number of turns in circular wire, N = 1

Consider R be the radius of the circular wire.

The magnetic field at the center of the current carrying circular wire is determine by the relation:

B=\frac{N\mu_{0} I}{2R}

Here μ₀ is vacuum permeability constant and its value is 4π x 10⁻⁷ Tm/A.

Substitute the suitable values in the above equation.

1.50\times10^{-4} =\frac{4\pi \times10^{-7}\times45 }{2R}

R = 0.18 m

4 0
2 years ago
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