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3 years ago

An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms–1

Physics

1 answer:

solniwko [45]3 years ago

6 0

Answer:

Initial velocity of the object, u = 5 m/s

Final velocity of the object, v = 8 m/s

Mass of the object, m = 100 kg

Time take by the object to accelerate, t = 6 s

Initial momentum = mu = 100 — 5 = 500 kg m sˆ’1

Final momentum = mv = 100 — 8 = 800 kg m sˆ’1

Force exerted on the object, F = mv – mu / t

= m (v-u) / t

= 800 – 500

= 300 / 6

= 50 N

Initial momentum of the object is 500 kg m sˆ’1.

Final momentum of the object is 800 kg m sˆ’1.

Force exerted on the object is 50 N.

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Use the graph below to answer the following question: if average acceleration is calculated using the equation, “ change in velo

sergiy2304 [10]

Answer:

$a=9\ cm/s^2$

Explanation:

<u>Average Acceleration </u>

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Note: <em>We cannot see if the time axis is numbered in increments of 1 second, and we'll assume that. </em>

When $t_2=4\ sec$ , the graph shows a value of $v_2=36\ cm/s$

When $t_1=0\ sec$ , the object is at rest, $v_1=0$

We compute the average acceleration as

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3 years ago

A space station has a large ring-like component that rotates to simulategravity for the crew. This ring has a massM= 2.1×105kg a

ivann1987 [24]

Answer:

Each thruster has to applied a force of 294.5N in tangential direction

Explanation:

Mass of the ring ,M =2.1×105kg

Mass of the ship ,m = 3.5×104kg

Radius of the ring R = 86.0 m

distance of ship from center of the ring, r =31.0 m

Let force applied by each thruster be F

Time taken to reach gravity ,t = 3hrs = 3600× 3 =10800sec

The movement of ring make the object kept at the edge feel a force of centrifuge in outward direction.

Centrifugal force = weight of the object on earth

Assume the ring is moving with angular speed ω

Centripetalforce of the object kept at ring

m₁R ω²=m₁g (m₁=mas of object)

⇒Rω² = g

⇒ω = √g/R

The ring start from 0 angular speed with constant angular acceleration

Let the constant angular acceleration be ∝

∝ = ω / t

(ω = √g/R)

∝ = $\frac{1}{t } \sqrt{\frac{g}{R} } }$

Consider Torque on the ring and ship system

T = FR + FR = 2FR

Moment of inertia of ring ship system

I = I(ring)+I(ship)+I(ship)

= MR² + mr² + mr² = MR² + 2mr²

angular acceleration of the ring ship system

∝ = $\frac{2FR}{MR^2 + 2mr^2}$

Now we have ,

∝ = $\frac{1}{t} \sqrt{\frac{g}{R} }$ , ∝ = $\frac{2FR}{MR^2+2mr^2}$

equating both values

We have,

$F = \frac{MR^2+2mr^2}{2Rt} \sqrt{\frac{g}{R} }$

where,

m = 2.1 × 10⁵kg, R = 86m, m = 3.5 × 10₄kg,

r = 31m, g = 9.8m/s² , t = 10800sec

$F = \frac{(2.1 \times 10^5 \times86^2)+(2\times3.5v10^4\times31^2) }{2\times86\times10800} \times\sqrt{\frac{9.8}{86} } \\\\F = 294.47N\\\approx294.5N$

Each thruster has to applied a force of 294.5N in tangential direction

3 0

4 years ago

You just calibrated a constant volume gas thermometer. The pressure of the gas inside the thermometer is 294.0 kPa when the ther

Travka [436]

Answer: 361° C

Explanation:

Given

Initial pressure of the gas, P1 = 294 kPa

Final pressure of the gas, P2 = 500 kPa

Initial temperature of the gas, T1 = 100° C = 100 + 273 K = 373 K

Final temperature of the gas, T2 = ?

Let us assume that the gas is an ideal gas, then we use the equation below to solve

T2/T1 = P2/P1

T2 = T1 * (P2/P1)

T2 = (100 + 273) * (500 / 294)

T2 = 373 * (500 / 294)

T2 = 373 * 1.7

T2 = 634 K

T2 = 634 K - 273 K = 361° C

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