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Bad White [126]
1 year ago
7

Two horses, Thunder and Misty, are accelerating a wagon 1.3 m/s2. The force of friction is 75 N. Thunder is pulling with a force

of 1,000 N, while Misty is pulling with a force of 800 N. The force of gravity is 14,700 N and the normal force is 14,700 N. What is the mass of the wagon? Round the answer to the nearest whole number. kg
Physics
2 answers:
sasho [114]1 year ago
7 0
Assumption both thunder and misty are pulling in same direction,
Net force= 1000N+800N-75N=1725N
Mass of wagon = 1725N/1.3ms^-2 = 1327kg
umka2103 [35]1 year ago
6 0

Answer: 1327 kg

Explanation:

First of all, we can notice that the forces along the vertical direction (gravity and normal force) are equal and opposite, so they cancel each other and we can ignore them.

The relationship between the net force on the horizontal direction and the acceleration of the wagon is given by Newton's second law:

\sum F = ma

where

\sum F is the net force on the wagon

m is the mass of the wagon

a is the acceleration

In this problem, the net force on the wagon is:

\sum F=1,000 N+800 N-75 N=1,725 N

the mass is unkown

and the acceleration is a=1.3 m/s^2

So we can re-arrange the equation above to find the mass of the wagon:

m=\frac{\sum F}{a}=\frac{1725 N}{1.3 m/s^2}=1327 kg

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A system of two paint buckets connected by a lightweight rope is released from rest with the 12.0-kg bucket 2.00 m above the flo
NISA [10]

Explanation:

The given data is as follows.

    Mass of small bucket (m) = 4 kg

    Mass of big bucket (M) = 12 kg

    Initial velocity (v_{o}) = 0 m/s

    Final velocity (v_{f}) = ?

  Height H_{o} = h_{f} = 2 m

and,    H_{f} = h_{o} = 0 m

Now, according to the law of conservation of energy

         starting conditions = final conditions

  \frac{1}{2}MV^{2}_{o} + Mgh_{o} + \frac{1}{2}mv^{2}_{o} + mgh_{o} = \frac{1}{2}MV^{2}_{f} + Mgh_{f} + \frac{1}{2}mv^{2}_{f} + mgh_{f}

     \frac{1}{2}(12)(0)^{2} + (12)(9.81)(2) + \frac{1}{2}(4)(0)^{2} + (4)(9.81)(0) = \frac{1}{2}(12)V^{2}_{f} + (12)(9.81)(0) + \frac{1}{2}(4)V^{2}_{f} + (4)(9.81)(2)

                 235.44 = 8V^{2}_{f} + 78.48

                V_{f} = 4.43 m/s

Thus, we can conclude that the speed with which this bucket strikes the floor is 4.43 m/s.

3 0
1 year ago
In 2014, the Rosetta space probe reached the comet Churyumov Gerasimenko. Although the comet's core is actually far from spheric
Viktor [21]

To solve this problem we will apply the concepts related to gravity according to the Newtonian definitions. From finding this value we will use the linear motion kinematic equations to find the time. Our values are

Comet mass M = 1.0*10^{13} kg

Radius r = 1.6km = 1600 m

Rock was dropped from a height 'h' from surface = 1m

The relation for acceleration due to gravity of a body of mass 'm' with radius 'r' is

g = \frac{GM}{R^2}

Where G means gravitational universal constant and M the mass of the planet

g = \frac{(6.67408*10^{-11})(1*10^{13})}{1600^2}

g = 2.607*10^{-4} m/s^2

Now calculate the value of the time

h = \frac{1}{2} gt^2

t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2(1)}{2.607*10^{-4}}}

t = 87.58s

The time taken for the rock to reach the surface is t = 87.58s

8 0
1 year ago
A block moves at 5 m/s in the positive x direction and hits an identical block, initially at rest. A small amount of gunpowder h
Anestetic [448]

Answer:

Speed of 1.83 m/s and 6.83 m/s

Explanation:

From the principle of conservation of momentum

mv_o=m(v_1 + v_2) where m is the mass, v_o is the initial speed before impact, v_1 and v_2 are velocity of the impacting object after collision and velocity after impact of the originally constant object

5m=m(v_1 +v_2)

Therefore v_1+v_2=5

After collision, kinetic energy doubles hence

2m*(0.5mv_o)=0.5m(v_1^{2}+v_2^{2})

2v_o^{2}=v_1^{2} + v_2^{2}

Substituting 5 m/s for v_o then

2*(5^{2})= v_1^{2} + v_2^{2}

50= v_1^{2} + v_2^{2}

Also, it’s known that v_1+v_2=5 hence v_1=5-v_2

50=(5-v_2)^{2}+ v_2^{2}

50=25+v_2^{2}-10v_2+v_2^{2}

2v_2^{2}-10v_2-25=0

Solving the equation using quadratic formula where a=2, b=-10 and c=-25 then v_2=6.83 m/s

Substituting, v_1=-1.83 m/s

Therefore, the blocks move at a speed of 1.83 m/s and 6.83 m/s

6 0
1 year ago
A system contains a perfectly elastic spring, with an unstretched length of 20 cm and a spring constant of 4 N/cm.
mote1985 [20]

Answer:

a) When its length is 23 cm, the elastic potential energy of the spring is

0.18 J

b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J

Explanation:

Hi there!

a) The elastic potential energy (EPE) is calculated using the following equation:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = stretched lenght.

Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).

First, let´s convert the spring constant units into N/m:

4 N/cm · 100 cm/m = 400 N/m

EPE = 1/2 · 400 N/m · (0.03 m)²

EPE = 0.18 J

When its length is 23 cm, the elastic potential energy of the spring is 0.18 J

b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:

EPE = 1/2 · 400 N/m · (0.06 m)²

EPE = 0.72 J

When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J

7 0
2 years ago
A square loop of wire with initial side length 10 cm is placed in a magnetic field of strength 1 T. The field is parallel to the
Fofino [41]

Answer:

2 x 10⁻³ volts

Explanation:

B = magnetic of magnetic field parallel to the axis of loop = 1 T

\frac{dA}{dt} = rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²

θ = Angle of the magnetic field with the area vector = 0

E = emf induced in the loop

Induced emf is given as

E = B \frac{dA}{dt}

E = (1) (20 x 10⁻⁴ )

E = 2 x 10⁻³ volts

E = 2 mV

7 0
2 years ago
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