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3 years ago

What is the mass of a crate if a force of 200 N causes it to accelerate at 8 m/s2? (Formula: F=ma)

Physics

2 answers:

AfilCa [17]3 years ago

8 0

Answer:

Explanation:

Got it right on edg

Ugo [173]3 years ago

4 1

25 kg

Explanation:

we can calculate the mass of the crate by using the Newton's second law:

$F=ma$

where

F is the force acting on the crate

m is the mass

a is the acceleration

In this problem, $F=200 N$ and $a=8 m/s^2$ , so we can re-arrange the equation above to calculate the mass of the crate:

$m=\frac{F}{a}=\frac{200 N}{8 m/s^2}=25 kg$

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3 years ago

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Three carts of masses 4.0 kg, 10kg, and 3.0 kg move on a frictionless track with speeds of v1 = 5.0m/s, v2=3.0m/s, and v3=-3.6 m

gogolik [260]

2.24 m/s is the calculated velocity.

Initial velocity (u) squared plus two times the acceleration (a) times the displacement equals final velocity (v) squared (s). Final velocity (v) is equal to the square root of initial velocity (u) squared plus two times the acceleration (a) times displacement when v is the variable being solved for (s).

The cart's masses and speeds are known.

M1 = 4.00 kg, M2 = 10.0 kg, M3 = 3.00 kg, etc.

v1 = 5.00 m/s = 5.00 m/s, v2 = 3.00 m/s = 3.00 m/s, v3 = -4.00 m/s = 4.00 m/s, and m1v1+m2v2+m3v3 = (m1+m2+m3) v=d frac m 1v 1+m 2v 2+m 3v 3, where (m1+m2 + m3) is the product of (v1 v 1+m2v2+m3v3).

"m 1+m 2+m 3" is equivalent to "m 1+m2+m3/m1v 1+m2v2 +m3v3"

the three carts' final velocities are calculated as follows: v=d frac

{4.00kg\sdot5.00m/s+10.0kg\sdot3.00m/s-3.00kg\sdot4.00m/s} 4.24m/s = 4.50kg+10.0kg+3.00kg vs. 4.50kg+10.0kg+3.00kg

5.00m/s/4.00kg/5.00m/s+10.0m/s/3.00m/s/4.00m/s =2.24m/s.

2.24 m/s is the calculated final velocity.

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2 years ago

A small rubber wheel is used to drive a large pottery wheel. The two wheels are mounted so that their circular edges touch. The

Goshia [24]

Answer:

$0.629\ \text{rad/s}^2$ counterclockwise

$9.98\ \text{s}$

Explanation:

$r_1$ = Small drive wheel radius = 2.2 cm

$\alpha_1$ = Angular acceleration of the small drive wheel = $8\ \text{rad/s}^2$

$r_2$ = Radius of pottery wheel = 28 cm

$\alpha_2$ = Angular acceleration of pottery wheel

As the linear acceleration of the system is conserved we have

$r_1\alpha_1=r_2\alpha_2\\\Rightarrow \alpha_2=\dfrac{r_1\alpha_1}{r_2}\\\Rightarrow \alpha_2=\dfrac{2.2\times 8}{28}\\\Rightarrow \alpha_2=0.629\ \text{rad/s}^2$

The angular acceleration of the pottery wheel is $0.629\ \text{rad/s}^2$ .

The rubber drive wheel is rotating in clockwise direction so the pottery wheel will rotate counterclockwise.

$\omega_i$ = Initial angular velocity = 0

$\omega_f$ = Final angular velocity = $60\ \text{rpm}\times \dfrac{2\pi}{60}=6.28\ \text{rad/s}$

t = Time taken

From the kinematic equations of linear motion we have

$\omega_f=\omega_i+\alpha_2t\\\Rightarrow t=\dfrac{\omega_f-\omega_i}{\alpha_2}\\\Rightarrow t=\dfrac{6.28-0}{0.629}\\\Rightarrow t=9.98\ \text{s}$

The time it takes the pottery wheel to reach the required speed is $9.98\ \text{s}$

4 0

3 years ago

Which of the following is an example of work?

EastWind [94]

Answer:

it's D

Explanation:

6 0

3 years ago

A car on a level road turns a quarter circle ccw. You learned in Chapter 8 that static friction causes the centripetal accelerat

pav-90 [236]

Answer:

Zero

Explanation:

As we know that the force and the motion direction should always be perpendicular to each other due to which the work is done by static friction be zero

Therefore

F.dcos(theta) = F.d cos(90) = 0

Hence, the work done by static friction is zero

Therefore the same is to be considered

3 0

3 years ago