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mel-nik [20]
3 years ago
14

If fnet equals fg, what is the magnitude and direction of the acceleration of the object? Assume that the object is on Earth's s

urface.
Physics
1 answer:
suter [353]3 years ago
3 0
You're describing free-fall. The acceleration is 9.8 meters per second-squared downward (toward the center of the Earth).
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If the accepted value of a wave is 121 m/s, who has the most accurate method of measuring the speed of a wave?
qaws [65]
The answer would be erin out of all of them thank me later :)
5 0
3 years ago
A plane travels down a runway 2750 m before it lifts off at an angle of 37 degrees from the horizontal. The plane has traveled 1
ss7ja [257]

Answer: 4.236km

Explanation:

Let's define the point (x, y) as:

x = horizontal distance moved.

y = vertical distance moved.

If the plane starts in the point (0, 0) then:

"A plane travels down a runway 2750 m before it lifts off..."

At this time, the position will be:

P = (0 + 2750m, 0) = (2750m, 0).

"it lifts off at an angle of 37 degrees from the horizontal. The plane has traveled 1.8 km since its wheels left the ground."

In this case, as the angle is measured from the horizontal, the components will be:

x = 1.8km*cos(37°) = 1.438km

y = 1.8km*sin(37°) =  1.083 km

Then the new position is:

P = (2750m + 1.438 km, 0 + 1.083 km)

Let's write it using the same units for all the quantities:

we know that

1km = 1000m

Then:

2750m = (2750/1000) km = 2.750 km.

Then we can write the new position as:

P = (2.750 km + 1.438km, 1.083km) = (4.188km, 1.083km)

Now, we define the displacement as the distance between the final position and the initial position.

The distance between two points (a, b) and (c, d) is:

D = √( (a  c)^2 + (b - d)^2)

In this case the points are:

(0, 0) for the initial position

(4.188km, 1.083km) for the final position.

And the displacement will be:

D = √( (4.188km - 0)^2 + (1.083 - 0)^2) = 4.236km

5 0
3 years ago
A 5.0 kg object moving at 5.0 m/s. KE = mv2 times 1/2
steposvetlana [31]

Answer: KE = 62.5J

Explanation:

Given that

Mass of object = 5kg

kinetic energy KE = ?

velocity of object = 5m/s

Since kinetic energy is the energy possessed by a moving object, and it depends on the mass (m) of the object and the velocity (v) by which it moves. Therefore, the object has kinetic energy.

i.e K.E = 1/2mv^2

KE = 1/2 x 5kg x (5m/s)^2

KE = 0.5 x 5 x 25

KE = 62.5J

Thus, the object has 62.5 joules of kinetic energy.

5 0
3 years ago
A car is traveling south is 200 kilometers from it’s starting point after 2 hours. What is the average velocity of the car
Lesechka [4]

Answer:

100

Explanation:

take note that v=d/t (velocity is distance over(divided by) time, so in this case it would be 200 (distance) divided by 2 (time) = 100

6 0
3 years ago
A sample of n2 gas occupies a volume of 746 ml at stp. What volume would n2 gas occupy at 155 ◦c at a pressure of 368 torr?
musickatia [10]

Answer:

2.41 L

Explanation:

We can solve the problem by using the ideal gas equation, which can be rewritten as:

\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}

where we have:

p_1 = 1.01\cdot 10^5 Pa (initial pressure is stp pressure)

V_1 = 746 mL = 0.746 L = 7.46\cdot 10^{-4}m^3 is the initial volume

T_1 = 0^{\circ}=273 K is the initial temperature (stp temperature)

p_2 = 368 torr = 4.9\cdot 10^4 Pa is the final pressure

V_2 = ? is the final volume

T=155^{\circ}=428 K is the final temperature

By substituting the numbers inside the formula and solving for V2, we find the final volume:

V_2 = \frac{p_1 V_1 T_2}{T_1 p_2}=\frac{(1.01\cdot 10^5 Pa)(7.46\cdot 10^{-4} m^3)(428 K)}{(273 K)(4.9\cdot 10^4 Pa)}=2.41\cdot 10^{-3} m^3

which corresponds to 2.41 L.

7 0
3 years ago
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