3 years ago

Can an object have increasing speed while its acceleration is decreasing?

Physics

1 answer:

alexdok [17]3 years ago

6 0

The best option is C. This is due to friction.

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While running around the track at school, Milt notices that he runs due East in the 100m homestretch and due West on the 100m ba

Sergeeva-Olga [200]

They have equal magnitudes and opposite directions.

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3 years ago

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The position (in radians) of a car traveling around a curve is described by Θ (t) = t 3 - 2t 2 - 4t + 10 where t (in seconds). W

FromTheMoon [43]

Answer: $\alpha =30-4=26 rad/s^2$

Explanation:

Given

Position of a car is given by

$\theta =t^3-2t^2-4t+10$

and angular speed is $\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega$

angular acceleration is

$\frac{\mathrm{d^2} \theta }{\mathrm{d} t^2}=\alpha$

$\alpha =6t-4$

at $t=5 s$

$\alpha =30-4=26 rad/s^2$

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3 years ago

What is the acceleration of a 50 kg object pushed with a net force of 500 newtons?

vodomira [7]

Using the formula F=ma
500N=50kg (a)
a= 10 m/s^2

3 0

3 years ago

Determine the acceleration due to gravity for low Earth orbit (LEO) given: MEarth = 6.00 x 1024 kg, rEarth = 6.40 x 106 m, G = 6

Nana76 [90]

Answer:

The answer to the question is as follows

The acceleration due to gravity for low for orbit is 9.231 m/s²

Explanation:

The gravitational force is given as

$F_{G}= \frac{Gm_{1} m_{2}}{r^{2} }$

Where $F_{G}$ = Gravitational force

G = Gravitational constant = 6.67×10⁻¹¹ $\frac{Nm^{2} }{kg^{2} }$

m₁ = mEarth = mass of Earth = 6×10²⁴ kg

m₂ = The other mass which is acted upon by $F_{G}$ and = 1 kg

rEarth = The distance between the two masses = 6.40 x 10⁶ m

therefore at a height of 400 km above the erth we have

r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m

and $F_{G}$ = $\frac{6.67*10^{-11} *6.40*10^{24} *1}{(6.8*10^{6})^{2} }$ = 9.231 N

Therefore the acceleration due to gravity = $F_{G}$ /mass

9.231/1 or 9.231 m/s²

Therefore the acceleration due to gravity at 400 kn above the Earth's surface is 9.231 m/s²

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3 years ago

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How fast should a moving clock travel if it is to be observed by a stationary observer as running at one-half its normal rate?A)

Elan Coil [88]

Answer:

Option (D) is correct.

Explanation:

Let the speed is v.

$\Delta t = \gamma \Delta t'\\\\\Delta t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\times \frac{\Delta t}{2}\\\\\sqrt{1-\frac{v^2}{c^2}} =\frac{1}{2}\\\\1-\frac{v^2}{c^2}=\frac{1}{4}\\\\\frac{3}{4}c^2 = v^2\\\\v = 0.87 c$

Option (D) is correct.

4 0

3 years ago