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marissa [1.9K]
3 years ago
15

a particle of mass 1.3kg is sliding down a frictionless slope inclined at 30 to the horizontal. the acceleration of the particle

down the slope is
Physics
1 answer:
OverLord2011 [107]3 years ago
4 0

Answer:

4.9\ m/sec^2

Explanation:

The computation of acceleration of the particle down the slope is shown below:-

data provided in the question

Particle of mass = 1.3 kg i,e sliding down

Inclined = 30 to the horizontal

based on the above information

Force is given by

N = mg\ cos \theta ............ 1

and sliding force is given by

F = mg\ sin\alpha

a = g(sin\ 30^{\circ})

= 9.8\times \frac{1}{2} m/sec^2

= = 4.9\ m/sec^2

Hence, the acceleration of the particle down the slope is 4.9 m/sec^2

You might be interested in
A professor sits at rest on a stool that can rotate without friction. The rotational inertia of the professor-stool system is 4.
Anestetic [448]

Answer:

\omega=0.37 [rad/s]  

Explanation:

We can use the conservation of the angular momentum.

L=mvR

I\omega=mvR

Now the Inertia is I(professor_stool) plus mR², that is the momentum inertia of a hoop about central axis.

So we will have:

(I_{proffesor - stool}+mR^{2})\omega=mvR

Now, we just need to solve it for ω.

\omega=\frac{mvR}{I_{proffesor-stool}+mR^{2}}

\omega=\frac{1.5*2.7*0.4}{4.1+1.5*0.4^{2}}      

\omega=0.37 [rad/s]  

I hope it helps you!

5 0
3 years ago
An observer sees a flower pot sail up and then back down past a window 2.45 m high. If the total time the pot is in the sight is
Fynjy0 [20]

Answer: 2.55meter

Explanation: Using the second equation of motion.

S{hieght} = U*t + {g*t²}/2

Where U is initial velocity =0m/s

g is acceleration due to gravity 10m/s²

t is time 1secs

So we have,

hieght = 0 + {g*t²}/2

hieght = {10*(1)²}/2

Total hieght travelled is 10/2

Which is 5 meter.

But we are asked to find the hieght above the window which as a hieght of 2.45meter.

So,

hieght above window would be

{5 - 2.45}meter

Which is 2.55 meter.

8 0
3 years ago
Need help asap; put you as brainliest​
Vera_Pavlovna [14]

Answer:

unmmmmmmmm I think the answerA

8 0
3 years ago
A conducting loop has an area of 0.065 m2 and is positioned such that a uniform magnetic field is perpendicular to the plane of
aev [14]

Answer:

initial magnetic field  1.306 T

Explanation:

We have given area of the conducting loop A=0.065m^2

Emf induced = 1.2 volt

Initial magnetic field B = 0.3 T

Time dt = 0.087 sec

We know that induced emf is given by e=\frac{d\Phi }{dt}=-A\frac{db}{dt}

1.2=0.065\times \frac{db}{0.087}

db=1.606T

So initial magnetic field = 1.606-0.3= 1.306 T

5 0
3 years ago
a spring gun initially compressed 2cm fires a 0.01kg dart straight up into the air. if the dart reaches a height it 5.5m determi
Vikki [24]

Answer:

2697.75N/m

Explanation:

Step one

This problem bothers on energy stored in a spring.

Step two

Given data

Compression x= 2cm

To meter = 2/100= 0.02m

Mass m= 0.01kg

Height h= 5.5m

K=?

Let us assume g= 9.81m/s²

Step three

According to the principle of conservation of energy

We know that the the energy stored in a spring is

E= 1/2kx²

1/2kx²= mgh

Making k subject of formula we have

kx²= 2mgh

k= 2mgh/x²

k= (2*0.01*9.81*5.5)/0.02²

k= 1.0791/0.0004

k= 2697.75N/m

Hence the spring constant k is 2697.75N/m

7 0
3 years ago
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