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schepotkina [342]
3 years ago
11

Hey I need help on this please???

Physics
1 answer:
Agata [3.3K]3 years ago
6 0

Answer: it's c or 4.0 m/s

Explanation:

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A box with a mass of 20 kilograms is sliding down a ramp tilted at an angle θ of 30° to the horizontal. Determine the accelerati
Bumek [7]

The acceleration of the box as it slides down the ramp is 4.9 m/s².

<h3>What is the acceleration of the box down the incline?</h3>

The acceleration of the box down the incline is determined by applying Newton's second law of motion as shown below;

F - Ff = ma

where;

  • F is the parallel force on the box
  • Ff is the frictional force on box = 0
  • m is the mass of the box
  • a is the acceleration of the box

F - 0 = ma

F = ma

mg sinθ = ma

g sinθ = a

where;

  • g is acceleration due to gravity
  • θ is the angle of inclination of the incline

a = 9.8 x sin(30)

a = 4.9 m/s²

Learn more about acceleration here: brainly.com/question/14344386

#SPJ1

8 0
1 year ago
HELPPPPP MEEEEE PLEASE I NEED TO SUBMIT IN LESS THAN 10 MINSS
DerKrebs [107]

Answer:

X = 2146.05 m

Explanation:

We need to understand first what is the value we need to calculate here. In this case, we want to know how far from the starting point the package should be released. This is the distance.

We also know that the plane is flying a certain height with an specific speed. And the distance we need to calculate is the distance in X with the following expression:

X = Vt   (1)

However we do not know the time that this distance is covered. This time can be determined because we know the height of the plain. This time is referred to the time of flight. And the time of flight can be calculated with the following expression:

t = √2h/g   (2)

Where g is gravity acceleration which is 9.8 m/s². Replacing the data into the expression we have:

t = √(2*2500)/9.8

t = 22.59 s

Now replacing into (1) we have:

X = 95 * 22.59

<h2>X = 2146.05 m</h2>

This is the distance where the package should be released.

Hope this helps

6 0
3 years ago
The complete combustion of a small wooden match produces approximately 512 cal of heat. How many kilojoules are produced? Expres
k0ka [10]

We have to covert 512 cal of heat in kilo joules.

As, 1 cal = 0.004184 kJ = 4.184 joules.

Therefore,

512 \ cal = 512 \times 4.184 \ J = 2142.208 \ J \\\\\ =  2.142 \times 10 ^3 J = 2.142 \ kJ

Thus, combustion of a small wooden match produces approximately ( in kilo joules ) is 2.142 kJ .

5 0
3 years ago
Two 1.0 kg masses are 4.0 m apart on a frictionless table. Each has 1.0μC of charge. Part A What is the magnitude of the electri
9966 [12]

Force between two charges is given by

F =\frac{kq_1q_2}{r^2}

F =\frac{9*10^9* 1* 10^{-6}* 1 * 10^{-6}}{4^2}

F = 5.625 * 10^{-4} N

Now in order to find the acceleration of each mass

we can use

F = ma

5.625 * 10^{-4} = 1 * a

a= 5.625 * 10^{-4} m/s^2

8 0
4 years ago
Read 2 more answers
A 525 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satel
Dafna1 [17]

Answer:

1.69\cdot 10^{10}J

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

E=-G\frac{mM}{2r}

where

G is the gravitational constant

m = 525 kg is the mass of the satellite

M=5.98\cdot 10^{24}kg is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m

So the initial total energy is

E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J

When the satellite hits the ground, it is now on Earth's surface, so

r=R=6370 km=6.37\cdot 10^6 m

so its gravitational potential energy is

U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J

And since it hits the ground with speed

v=1.90 km/s = 1900 m/s

it also has kinetic energy:

K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J

So the total energy when the satellite hits the ground is

E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:

\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J

8 0
3 years ago
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