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3 years ago

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An object pulled to the right by two forces has an acceleration of 2.5m/s2. The free-body diagram shows the forces acting on the

object. What is the weight of the object?

1 answer:

Vanyuwa [196]3 years ago

8 0

Answer:

490 N is the correct answer.

Explanation:

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How do I calculate the tension in the horizontal string?

matrenka [14]

ANSWER

T₂ = 10.19N

EXPLANATION

Given:

• The mass of the ball, m = 1.8kg

First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,

In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

$\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}$

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

$\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}$

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

$T_1\cos 30\degree-m\cdot g=0$

Solve for T₁,

$T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N$

Now, we use the second equation to find the tension in the horizontal string,

$T_2-T_1\sin 30\degree=0$

Solve for T₂,

$T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N$

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

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1 year ago

A boy jogs 250 meters in 110s what is the average speed

Artyom0805 [142]

The rule to get the average speed is as follows:
average speed = average distance / average time

We are given that:
distance = 250 m
time = 110 sec

Substitute with the givens in the above equation to get the average speed as follows:
average speed = 250/110 = 25/11 meters/sec

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3 years ago

Solids diffuse because the particles cannot move.

Nastasia [14]

Solids cannot diffuse.

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3 years ago

Read 2 more answers

A pen rolls off a 0.55–meter high table with an initial horizontal velocity of 1.2 meters/second. At what horizontal distance fr

NARA [144]

To find the horizontal distance multiple the horizontal velocity by the time. Since there is no given time it must be calculated using kinematic equation.

Y=Yo+Voyt+1/2at^2
0=.55+0+1/2(-9.8)t^2
-.55=-4.9t^2
sqrt(.55/4.9)=t
t=0.335 seconds

Horizontal distance
=0.335s*1.2m/s
=0.402 meters

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3 years ago

Electrical energy can be transformed into other types of energy. We often experience this transformation of energy in our everyd

vitfil [10]

Well I’m not sure because you don’t have anything listed

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3 years ago