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2 years ago

If the acceleration of an object increases, what effect does this have on the Force and Mass?

Physics

2 answers:

Hitman42 [59]2 years ago

8 0

First of all mass stays the same that never changes.

However when using the force times acceleration (The net force on an object is equal to the mass of the object multiplied by the acceleration of the object. Or some simply say: Force equals mass times acceleration.)

Answer: Force Increases and Mass remains the same.

Hope this helps!

Nutka1998 [239]2 years ago

5 0

Answer:

Force Increases and Mass remains the same.

Explanation:

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Piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note

mote1985 [20]

Answer:

Frequency = 658 Hz

Explanation:

The third harmonics of A is,

$f_{A}=(3)(440Hz)=1320Hz$

The second harmonics of E is,

$f_{E}=(2)(659Hz)=1318Hz$

The difference in the two frequencies is,

delta_f = 1320 Hz - 1318 Hz = 2 Hz

The beat frequency between the third harmonic of A and the second harmonic of E is,

delta_f = $3f_{A}-2f_{E}$

$f_{E}=\frac{3f_{A}-delta_f}{2}$

We have calculate the frequency of the E string when she hears four beats per second, then

delat_f = 4 Hz

$f_{E}=\frac{3(440Hz)-4Hz}{2}$

$f_{E}=658Hz$

Hope this helps!

5 0

3 years ago

Which wavelength of light is capable of penetrating the dust of a nebula?

ivann1987 [24]

When the dust is too thick to penetrate with visible light, such as the Nebula, Radio Waves are used to penetrate the dust. Longer radio waves can completely penetrate the thick cloud cover, allowing scientists to beam radar waves.

8 0

3 years ago

The layer of earth that has the lightest elements is the

Sonja [21]

If I remember correctly (from my studies long time ago) the layers are from the outer to the center:
SiAl : Silicon-Aluminum
SiMa : Silicon-Magnesium (although should be Mg)
NiFe : Nickel-Iron

The SiMa layer should have the lightest elements (Magnesium is lighter than Aluminum)

8 0

4 years ago

The basic barometer can be used as an altitude-measuring device in airplanes. The ground control reports a barometric reading of

kipiarov [429]

Answer:

Δh_air=714m

Explanation:

Given data

$P_{1}=753mmHg\\P_{2}=690mmHg\\ p_{air}=1.2kg/m^{3}\\ g=9.8m/s^{2}$

Solution

ΔP=P₁-P₂

=(ΔhHg)×pHg×g

=(Δh_air)× p_air ×g

Then

Δh_air=(pHg+ΔhHg)÷p_air

$=\frac{13600*(753-690)*10^{-3} }{1.2}\\ =714m$

Δh_air=714m

7 0

3 years ago

Apply the general results obtained in the full analysis of motion under the influence of a constant force in Section 2.5 to answ

zvonat [6]

Answer:

y(i) = h

v(y.i) = 0

Explanation:

See attachment for elaboration

3 0

3 years ago