All categories

2 years ago

A force acts on a 5kg object at rest. How fast will the object accelerate on a frictionless surface?

Physics

1 answer:

MakcuM [25]2 years ago

8 0

Answer: The answer is C.) 25 m/s^2.

Explanation: If you input 5 as s, you would have to use the exponent 2. This means that you have to multiply 5 by 5. 5 x 5= 25.

Edit: Also, because the surface is frictionless, it will make the object go faster too. Nothing can really slow it down unless something blocks it.

You might be interested in

Juno created a diagram to describe characteristics of freezing and deposition. Which labels belong in the areas marked X and Y?

DIA [1.3K]

the correct answer to this question is A

7 0

2 years ago

Read 2 more answers

A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of 3

Advocard [28]

Answer:

r = 5.07 m

Explanation:

given,

velocity of the man , v = 3.43 m/s

centripetal acceleration, a = 2.32 m/s²

magnitude of position of = ?

using centripetal acceleration formula

$a_c =\dfrac{v^2}{r}$

$2.32 =\dfrac{3.43^2}{r}$

$r =\dfrac{3.43^2}{2.32}$

r = 5.07 m

The magnitude of the position vector relative to rotational axis is equal to 5.07 m.

5 0

2 years ago

What is Newtons laws

pogonyaev

<h2>Law 1:</h2><h3>An object already in motion stays in motion, unless acted upon by a force.</h3><h3 /><h2>Law 2:</h2><h3> $f=ma$ </h3><h3>f = forces on an object</h3><h3>m = mass of that object</h3><h3>a = acceleration of that object</h3><h3 /><h2>Law 3:</h2><h3>Everything has an equal and opposite reaction.</h3><h3 /><h3>Hope this helps!</h3>

5 0

3 years ago

A force of 5N and a force of 8N act to the same point and are inclined at 45degree to each other. Find the magnitude and directi

Alex_Xolod [135]

Magnitude: 12.1 N.
Direction: 17.0° to the 8 N force.

<h3>Explanation</h3>

Refer to the diagram attached (created with GeoGebra). Consider the 5 N force in two directions: parallel to the 8 N force and normal to the 8 N force.

$\displaystyle F_{\text{1, Parallel}} = F_1 \cdot \cos{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}$ .
$\displaystyle F_{\text{1, Normal}} = F_1 \cdot \sin{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}$ .

The sum of forces on each direction will be the resultant force on that direction:

Resultant force parallel to the 8 N force: $(8 + \dfrac{5\sqrt{2}}{2})\;\text{N}$ .
Resultant force normal to the 8 N force: $\dfrac{5\sqrt{2}}{2}\;\text{N}$ .

Apply the Pythagorean Theorem to find the magnitude of the resultant force.

$\displaystyle \Sigma F = \sqrt{{(8 + \dfrac{5\sqrt{2}}{2})}^2 + {(\dfrac{5\sqrt{2}}{2})}^2} = 12.1\;\text{N}$ (3 sig. fig.).

The size of the angle between the resultant force and the 8 N force can be found from the tangent value of the angle. Tangent of the angle:

$\displaystyle \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}} = \dfrac{8 + \dfrac{5\sqrt{2}}{2}}{\dfrac{5\sqrt{2}}{2}} \approx 0.306491$ .

Find the size of the angle using inverse tangent:

$\displaystyle \arctan{ \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}}} = \arctan{0.306491} = 17.0\textdegree$ .

In other words, the resultant force is 17.0° relative to the 8 N force.

4 0

2 years ago

A piece of aluminum has a volume of 1.50 10-3 m3. the coefficient of volume expansion for aluminum is β = 69 ✕ 10-6 (°c)-1. the

Alex17521 [72]

Answer:

W = 3.12 J

Explanation:

Given the volume is 1.50*10^-3 m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:

β = 69*10^-6 (°C)^-1 V = 1.50*10^-3 m^3 ΔT = 298°C

So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):

$dV = \beta V_{0} dT\\dV = (69*10^{-6})( C)^{-1} * (1.50*10^{-3})m^{3} * (298)C\\dV = 3.0843*10^-5$

So ΔV = 3.0843*10^-5 m^3

Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa

To get work, multiply the air pressure and the volume change.

$W = P * dV = (1.01 * 10^5)Pa * (3.0843*10^{-5})m^3\\W = 3.115143J$

W = 3.12 J

Hope this helps!

4 0

3 years ago