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2 years ago

A force gives a 2.0 kg mass an acceleration of 5.0 m/s2 on a level surface. What is the force applied to the mass?

Physics

1 answer:

Verdich [7]2 years ago

8 0

So you can use the equation force = mass x acceleration to do 2 x 5 to get 10 N

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A vector has a component of 10 m in the x-direction, a component of 10 m in the y-direction, and a component of 5 m in the z dir

Kobotan [32]

The magnitude of this vector is 15

A vector is a quantity or phenomenon that has two independent properties: magnitude and direction. The term also denotes the mathematical or geometrical representation of such a quantity. Examples of vectors in nature are velocity, momentum, force, electromagnetic fields, and weight.

The magnitude of a vector formula is used to calculate the length for a given vector (say v) and is denoted as |v|. So basically, this quantity is the length between the initial point and endpoint of the vector.

Let vector be = a

component of vector in x direction = 10 i

component of vector in y direction = 10 j

component of vector in z direction = 5 z

vector a = 10 i + 10 j + 5 z

magnitude of vector a = |a| = $\sqrt{10^{2} +10^{2} + 5^{2} }$

= 15

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1 year ago

a stomp rocket takes 1.5 seconds to reach its maximum height what was the initial velocity and what was the maximum height ?

Zinaida [17]

1) Vf = Vo - gt; Vf = 0 => Vo = gt = 9.8m/s^2 * 1.5s = 14.7 m/s

2) d = Vo*t - gt^2 /2 = 14.7m/s*1.5 - 9.8m/s^2 * (1.5s)^2 / 2 = 11.02 m

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2 years ago

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A(n) ___ keeps a reaction from occurring by increasing the activation energy required for a reaction to occur. A. inhibitor B. c

skelet666 [1.2K]

Due to requirements i am typing this but the ans is a

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2 years ago

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Which has more thermal energy 1 g if aluminum or 1 g of gold and why

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2 years ago

A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpend

Sonja [21]

The time interval for which the proton remains in the field is -

Δt = $$\frac{\pi R}{40}$ .

We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.

We have to determine time interval during which the proton is in the field.

<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>

The magnitude of force on the charged particle moving in a uniform magnetic field is given by -

F = qvB sinθ

According to the question, we have -

Entering Velocity (v) = 20 i m/s

Magnetic field intensity (B) = 0.3 T

Leaving velocity (u) = - 20 j m/s

Now -

The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -

d = $$\frac{2\pi r}{4}$ = $$\frac{\pi R}{2}$

Since, the radius of circular path is not given, we will assume it R.

Therefore, time for which proton remained in the field is -

t = $$\frac{\pi R}{2v} = \frac{\pi R}{40}$

Hence, the time interval for which the proton remains in the field is -

Δt = $$\frac{\pi R}{40}$

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1 year ago