3 years ago

How much force is required to accelerate a 2 kg mass at 3 m/s^2?*

Physics

1 answer:

alexgriva [62]3 years ago

5 0

Answer: 6N

Explanation:

F = MA due to Newton's second law

M = 2

A = 3

plug in to get

F = 2 * 3 = 6N

You might be interested in

A physics student swings a tennis ball connected to a rope in a vertical circle with a constant speed of 6.29 m/s. The ball has

Alex777 [14]

Answer:

r = 0.5 m

Explanation:

First we find the angular speed of the ball by using its period:

ω = θ/t

For the time period:

ω = angular speed = ?

θ = angular displacement = 2π rad

t = time period = 0.5 s

Therefore,

ω = 2π rad/0.5 s

ω = 12.56 rad/s

Now, for the radius:

v = rω

r = v/ω

where,

v = linear speed = 6.29 m/s

r = radius = ?

r = (6.29 m/s)/(12.56 rad/s)

8 0

2 years ago

Calculate the volume of the metal required to make a hemisperical bowl with internal and external radii 8.4cm and 9.1cm respecti

ioda

External = R
Internal = r
Volume of hemisperical = 2/3 π(R³-r³)
V= 2/3 π(9.1³ - 8.4³)
V= 336.9 cm³

4 0

2 years ago

PLEASE HELP!! ITS DUE AT 11:59!!

DENIUS [597]

Answer:

14.1 po

sana makatulong

3 0

2 years ago

If an object is projected horizontally from a height of 5 m with an initial velocity of 7 m/s, what is the value of x0?

tatiyna

ANSWER IS D X0=9.8m/s^2

5 0

1 year ago

A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind

Marat540 [252]

Answer:

Part a)

$F_v = 4.28 N$

Part B)

$L = 1.02 m$

Part C)

$v = 1.25 m/s$

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

$Tcos\theta = mg$

$T sin\theta = F_v$

$\frac{F_v}{mg} = tan\theta$

$F_v = mg tan\theta$

$F_v = 1.2\times 9.81 (tan20)$

$F_v = 4.28 N$

Part B)

Here we can use energy theorem to find the distance that it will move

$-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2$

$(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2$

$(-3.5 + 2.54)L = - 0.98$

$L = 1.02 m$

Part C)

At terminal speed condition we know that

$F_v = mg$

$bv^2 = mg$

$2.5 v^2 = 3.9$

$v = 1.25 m/s$

7 0

2 years ago