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2 years ago

A sled starts from rest,

Physics

1 answer:

Scorpion4ik [409]2 years ago

6 0

Answer:

The acceleration of the sled is $4.6\ m/s^2$ .

Explanation:

It is given that,

Initial speed of sled is 0 because it was at rest.

It is placed at an angle of 28° on a frictionless hill.

We need to find the acceleration of the sled. It is placed at an incline. It means that the acceleration of the sled is given by :

$a=g\sin\theta\\\\a=9.8\times \sin(28)\\\\a=4.6\ m/s^2$

So, the acceleration of the sled is $4.6\ m/s^2$ .

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A block is pulled across a flat surface at a constant speed using a force of 50 newtons at an angle of 60 degrees above the hori

vladimir2022 [97]

The magnitude of the friction force is 25 N

Explanation:

To solve this problem, we just have to analyze the forces acting on the block along the horizontal direction. We have:

The horizontal component of the pulling force, $F cos \theta$ , where F = 50 N is the magnitude and $\theta=60^{\circ}$ is the angle between the direction of the force and the horizontal; this force acts in the forward direction
The force of friction, $F_f$ , acting in the backward direction

According to Newton's second law, the net force acting on the block in the horizontal direction must be equal to the product between the mass of the block and its acceleration:

$\sum F_x = ma_x$

where

m is the mass of the block

$a_x$ is the horizontal acceleration

However, the block is moving at constant speed, so the acceleration is zero:

$a_x = 0$

So the equation becomes

$\sum F_x = 0$ (1)

The net force here is given by

$\sum F_x = F cos \theta - F_f$ (2)

And so, by combining (1) and (2), we find the magnitude of the friction force:

$F cos \theta - F_f = 0\\F_f = F cos \theta = (50)(cos 60^{\circ})=25 N$

Learn more about force of friction:

brainly.com/question/6217246

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#LearnwithBrainly

4 0

2 years ago

I need help to figure out how to solve this problem and solve it!!!

wariber [46]

well it looks like the walk at a constant increasing pace then at a constant pace then increaseing pace then constant pace then they slow down then walk at a constant pace then walk at a constantly increasing pace

plz rate me brainliest

4 0

3 years ago

Read 2 more answers

"The drawing shows three layers of different materials, with air above and below the layers. The interfaces between the layers a

My name is Ann [436]

Answer:

Angle of incidence that entered material b= 63.1°

Angle of incidence between a and b = 55.9°

Explanation: Using the formular:

n1sintheta1= n2sintheta2

The light ray which enters material B will be

1.4Sin72.8° = 1.5Sin theta

1.3373= 1.5Sintheta

sintheta = 1.3373/1.5

Sin^-1 0.8916 = Theta

63.1 = theta

When the ray hits interface with material a

1.5Sin63.1 = 1.3 Sin theta

1.3374 = 1.3Sin theta

Sintheta= 1.3374/1.3

Sin theta = 1.0877

There will be total reflection off the boundary b c because sin theta exceeded 1 in value.

The equation should be

1.4sin63.1 = 1.4 sin theta

Sin theta=72.8°

When the ray hits air-c boundary:

1.4sin72.8=1.00sin theta

Sin theta=1.3374/1 =1.3374

There is total reflection.

In material a,the ray will:

1.3sin72.8° = 1.00sin theta

There will be total reflection when the ray hits a-b boundary.

1.3sin72.8= 1.5sintheta

Sin theta= 1.2419/ 1.5

Sin theta =0.8279

Theta= Sin^-10.8279= 55.88°

When ray hits c-air boundary

1.4sin63.1= 1.00sintheta

1.2485= sin theta = Toal reflection.

Therefore when the ray of light pass through the layers of material a, b and c the boundary with air on top and bottom will be total reflection.

7 0

3 years ago

Fill in the term that completes the statement. The north and south poles of a magnetic field produced by an electromagnet will s

Minchanka [31]

Answer:

when the direction of the Current changes.

Explanation:

Electromagnet refers to an iron ore wrapped around with a coil of wire, in presence of electric current. As it acts like a magnet, when current is passed through it.

The north & south poles of magnetic fields produced by such magnet, change with direction of current passed through it.

6 0

2 years ago

Read 2 more answers

What is the radius of a communications satellite’s orbital path that is in a uniform circular orbit around Earth and has a perio

pantera1 [17]

If the period of a satellite is T=24 h = 86400 s that means it is in geostationary orbit around Earth. That means that the force of gravity Fg and the centripetal force Fcp are equal:

Fg=Fcp

m*g=m*(v²/R),

where m is mass, v is the velocity of the satelite and R is the height of the satellite and g=G*(M/r²), where G=6.67*10^-11 m³ kg⁻¹ s⁻², M is the mass of the Earth and r is the distance from the satellite.

Masses cancel out and we have:

G*(M/r²)=v²/R, R=r so:

G*(M/r)=v²

r=G*(M/v²), since v=ωr it means v²=ω²r² and we plug it in,

r=G*(M/ω²r²),

r³=G*(M/ω²), ω=2π/T, it means ω²=4π²/T² and we plug that in:

r³=G*(M/(4π²/T²)), and finally we take the third root to get r:

r=∛{(G*M*T²)/(4π²)}=4.226*10^7 m= 42 260 km which is the height of a geostationary satellite.

3 0

2 years ago