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Bad White [126]
2 years 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]2 years 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]2 years 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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Aleks04 [339]

Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

\theta = Angle = 20°

y = -20 cm

Velocity components

u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s

u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s

Acceleration components

a_x=0

a_y=-9.81\ m/s^2

y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0

t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629

Time taken is 0.26088 seconds

x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m

The distance the beetle travels on the ground is 0.3677181864 m

6 0
2 years ago
A vertical spring of constant k = 400 N/m hangs at rest. When a 2 kg mass is attached to it, and it is released, the spring exte
Viefleur [7K]

Answer:

4.9 cm

Explanation:

From Hook's Law,

F = ke......................... Equation 1

Where F= force, e = extension, k = spring constant.

Note: the Force acting on the the spring is the weight of the mass.

W = mg.

F = mg.................... Equation 2

Where m = mass, g = acceleration due to gravity

Substitute equation 2 into equation 1

mg = ke

make e the subject of the equation

e = mg/k............... Equation 3.

Given: m = 2 kg, g = 9.8 m/s², k = 400 N/m

e = (2×9.8)/400

e = 19.6/400

e = 0.049 m

e = 4.9 cm

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