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Scrat [10]
3 years ago
15

What is the correct formula for of the weight w of an object? assume that its mass is m, the magnitude of its acceleration is a,

and the local acceleration due to gravity is g?
Physics
2 answers:
Vsevolod [243]3 years ago
6 0
Weight = Mass × Acceleration
Step2247 [10]3 years ago
4 0
W = m • g

Weight = mass x local gravity.

The object's acceleration doesn't matter.
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A.) If its booster rockets accelerate the space shuttle at 15m/s2, how high will it be one minute after launch?
poizon [28]

Answer:

27,000 m

450 m/s

Explanation:

Assuming the initial velocity is 0 m/s:

v₀ = 0 m/s

a = 15 m/s²

t = 60 s

A) Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (60 s) + ½ (15 m/s²) (60 s)²

Δy = 27,000 m

B) Find: v_avg

v_avg = Δy / t

v_avg = 27,000 m / 60 s

v_avg = 450 m/s

5 0
3 years ago
Explain why a ball thrown in space could keep moving forever, while a ball thrown here on Earth will come to a stop.
Dmitriy789 [7]

Answer:

In space there is no air resistance. on earth there is

in space there is no opposite forces acting on stopping the ball, so if you throw it it will go on forever.

on earth there is air resistance and gravity, this will pull the ball towards the ground and slow it down.

4 0
2 years ago
Describe the process you used to build a model
kherson [118]
Step 1 : Get your supply list together

Step 2 : Pick what model you want to do

Step 3 : Ask for a partner 

Step 4 : Complete  the model and take your time.

Step 5 : Read the directions carefully 
6 0
3 years ago
Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as
MAXImum [283]

Answer:

108 km

Explanation:

The conversion factor between meters and feet is

1 m = 3.28 ft

So the second altitude, written in feet, can be rewritten in meters as

h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m

or in kilometers,

h_2 = 47 km

the first altitude in kilometers is

h_1 = 155 km

so the difference between the two altitudes is

\Delta h = 155 km - 47 km = 108 km

8 0
3 years ago
For a photoelectric tube, calculate the voltage which will be just sufficient to stop electrons emitted by the sodium photo-plat
Talja [164]

Answer:

1.11 V

Explanation:

Given that the Einstein photoelectric equation states that;

KE = E - Wo

E = energy of incident photon

Wo= work function of the metal

E = hf = 6.64 x 10-34 * 6 x 1014

E = 39.84 * 10^-20 J or 3.98  * 10^-19 J

KE = 3.98  * 10^-19 J - 2.2 x 10-19J

KE = 1.78 *  10^-19J

We convert this value of KE to electron volts

KE = 1.78 *  10^-19J/1.6 x 10-19C

KE = 1.11 eV

Hence; 1.11 V will be just sufficient to stop electrons emitted by the sodium photo-plate reaching the collector plate.

8 0
3 years ago
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