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2 years ago

The forces in (Figure 1) are acting on a 1.0 kg object.What is ax , the x -component of the object's acceleration

Physics

1 answer:

a_sh-v [17]2 years ago

3 0

The x -component of the object's acceleration is 2 m/s².

<h3>What's the resultant force along x- direction?</h3>

Forces along x axis direction are as follows

4N along +x axis, so it's taken as +4 N
2N along -x axis , so it's taken as -2N.

Resultant force along x direction = 4N - 2N = 2 N which is along + ve x direction.

<h3>What's the acceleration along x axis direction?</h3>

As per Newton's second law, Force = mass × acceleration of the object
Force along x axis= mass × acceleration along x axis= 2N
Acceleration = 2/ mass = 2/1 = 2 m/s²

Thus, we can conclude that the acceleration along x axis is 2 m/s².

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: The forces in (Figure 1) are acting on a 1.0 kg object. What is ax, the x-component of the object's acceleration?

Learn more about the acceleration here:

brainly.com/question/460763

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irius, the brightest star in the sky, is 2.6 parsecs (8.6 light-years) from Earth, giving it a parallax of 0.379 arcseconds. Ano

Norma-Jean [14]

The actual distance of Regulus from Earth is 23.81 parsecs.

Given:

Parallax of Regulus, p = 0.042 arc seconds

Calculation:

When an observer changes their position, an apparent change in the object's position takes place. This change can be calculated using the angle ( or semi-angle) made by the observer and object i.e. the angle made between the two lines of observation from the object to the observer.

Thus from the relation of parallax of a celestial body we get:

S = 1/ tan p ≈ 1 / p

where S is the actual distance between the object and the observer

p is the parallax angle observed

Here for Regulus, we get:

S = 1 / p

= 1 / (0.042) [ 1 parsecs = 1 arcseconds ]

= 23.81 parsecs

We know that,

1 parsecs = 3.26 light-years = 206,000 AU

Converting the actual distance into light years we get:

23.81 parsecs = 23.81 × (3.26 light yrs) = 77.658 light-years

Therefore, the actual distance of Regulus from Earth is 23.81 parsecs which is 77.658 in light years.

Learn more about astronomical units here:

<u>brainly.com/question/16471213</u>

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6 0

1 year ago

If i click on a mouse does that show energy transfer?

Radda [10]

Answer:

Yes

Explanation:

You are using energy to click the mouse, and the energy moves from your fingers to the mouse clicker.

8 0

2 years ago

Read 2 more answers

The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have st

irina [24]

Complete Question:

The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 mm , while nonathletes' stretch only 32 mm . The spring constant for the tendon is the same for both groups, $31 {\rm {N}/{mm}}$ . What is the difference in maximum stored energy between the sprinters and the nonathlethes?

Answer:

$\triangle E = 12.79 J$

Explanation:

Sprinters' tendons stretch, $x_s = 43 mm = 0.043 m$

Non athletes' stretch, $x_n = 32 mm = 0.032 m$

Spring constant for the two groups, k = 31 N/mm = 3100 N/m

Maximum Energy stored in the sprinter, $E_s = 0.5kx_s^2$

Maximum energy stored in the non athletes, $E_m = 0.5kx_n^2$

Difference in maximum stored energy between the sprinters and the non-athlethes:

$\triangle E = E_s - E_n = 0.5k(x_s^2 - x_n^2)\\\triangle E = 0.5*3100* (0.043^2 - 0.032^2)\\\triangle E = 0.5*31000*0.000825\\\triangle E = 12.79 J$

4 0

3 years ago

if a man hits a golf ball (0.2 kg) which accelerates at a rate of 20 m/s squared. what amount of force acted on the ball?

kkurt [141]

F = ma

We have mass = 0.2kg

and acceleration = 20 m/s^2

So..

F = (0.2)(20)

F = 4 N

6 0

2 years ago

An inventor claims to have developed a power cycle having a thermal efficiency of 40%, while operating between hot and cold rese

lesya [120]

From Carnot's theorem, for any engine working between these two temperatures:

efficiency <= (1-tc/th) * 100

Given: tc = 300k (from question assuming it is not 5300 as it seems)

For a, th = 900k, efficiency = (1-300/900) = 70%

For b, th = 500k, efficiency = (1-300/500) = 40%

For c, th = 375k, efficiency = (1-300/375) = 20%

Hence in case of a and b, efficiency claimed is lesser than efficiency calculated, which is valid case and in case of c, however efficiency claimed is greater which is invalid.

7 0

2 years ago