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jarptica [38.1K]
3 years ago
12

A 225 kg block is pulled by two horizontal forces. The first force is 178 N at a 41.7-degree angle and the second is 259 N at a

108-degree angle. What is the x and y component of the total force acting on the block?
Physics
1 answer:
yawa3891 [41]3 years ago
7 0

Answer:

52.9 N, 364.7 N

Explanation:

First of all, we need to resolve both forces along the x- and y- direction. We have:

- Force A (178 N)

A_x = (178 N)(cos 41.7^{\circ})=132.9 N\\A_y = (178 N)(sin 41.7^{\circ})=118.4 N

- Force B (259 N)

B_x = (259 N)(cos 108^{\circ})=-80.0 N\\B_y = (259 N)(sin 108^{\circ})=246.3 N

So the x- and y- component of the total force acting on the block are:

R_x = A_x + B_x = 132.9 N - 80.0 N =52.9 N\\R_y = A_y + B_y = 118.4 N +246.3 N = 364.7 N

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EXPLANATION :

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The potential energy of a body due to its position from ground is known as gravitational potential energy.

The gravitational potential energy is calculated as -

                      Potential energy P.E = mgh

 Here, m is the mass of the body, and g is the acceleration due to gravity.

h stands for the height of the body from the ground.

We know that weight of a body is equal to the product of mass with acceleration due to gravity.

Hence, weight W = mg

Hence, potential energy is written as P.E = weight × height.

Hence, potential energy depends on the weight and height of the water.


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3 years ago
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A charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a th
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Answer:

x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}

Explanation:

Position of charge 3q is x = 0

position of charge -2q is x = a

so here we know that

when two charges are of opposite nature then the electric field will be zero on the line joining the two charges at the position near to smaller magnitude charge

So here the electric field will be zero if the field due to 3q is counterbalanced by field due to -2q

so here we can say

\frac{k(3q)}{(r+a)^2} + \frac{k(-2q)}{r^2} = 0

\frac{3}{(r + a)^2} = \frac{2}{r^2}

\frac{r+a}{r} = \sqrt{\frac{3}{2}}

1 + \frac{a}{r} =\sqrt{\frac{3}{2}}

\frac{a}{r} = \frac{\sqrt3 - \sqrt2}{\sqrt2}

so we will have

r = \frac{a\sqrt2}{\sqrt3 - \sqrt2}

so the x coordinate of this position is given as

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6 0
3 years ago
BRAINLIEST
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Answer:

Explanation:

1) acceleration is the change in velocity of a body with respect to time.

acceleration = velocity/time

Given

velocity = 139m/s

time = 20secs

acceleration = 139/20

acceleration = 6.95m/s²

Hence its average acceleration during the first 20 seconds of the launch is 6.95m/s²

2) Speed is the rate of change of distance with respect to time.

average speed = distance/time

Time = distance/speed

Time = 100/7823

Time = 0.013s

3) Using the equation of motion v² = u²+2as

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Answer:

73.67 m

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V(t) = V_0 + a * t,

Where y_0 its our initial height, V_0  our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

y(t) =  38 \frac{m}{s} * t - \frac{1}{2} g t^2

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0 m = 38 \frac{m}{s} - g * t

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t = 38 \frac{m}{s} / g

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Plugin this time on our first equation we find:

y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2

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