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4 years ago

12

A certain unbalanced force gives a 5kg object an acceleration of 15 m/s2. What would be the acceleration if the same force was a

pplied to 30 kg object ?

1 answer:

stira [4]4 years ago

7 0

Answer:

a2 = 2.5 m/s2

Explanation:

F1 = m1 a1 We use the same force so F1 = F2

= 5kg × 15m/s2 F2 = m2 a2

= 75N a2 is required

a2 = F2 / m2

= 75N / 30 kg

= 2.5 m/s2

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A 1000 Kg car approaches an intersection traveling north at 30 m/s . A 1250 Kg car approaches the same intersection traveling ea

Feliz [49]

Answer:

The final velocity has a magnitude of 25.44 m/s and is at 31.61° north of east.

Explanation:

Taking north direction as positive y axis and east direction as positive x axis .

Given:

Mass of first car is, $m_1=1000\ kg$

Initial velocity of first car is, $u_1=30\vec{j}\ m/s$

Mass of second car is, $m_2=1250\ kg$

Initial velocity of second car is,

Let the combined final velocity after collision be 'v' m/s with as components of final velocity along east and north directions respectively.

Now, as the net external force is zero, momentum is conserved for the two car system along the east and north directions.

Conserving momentum along the east direction, we have:

Initial momentum = Final momentum

$m_1u_{1x}+m_2u_{2x}=(m_1+m_2)v_x\\\\0+1250\times 39=(1000+1250)v_x\\\\v_x=\frac{48750}{2250}\\\\v_x=\frac{65}{3} m/s$

There is no component of initial velocity for first car in east direction, as it is moving in the north direction. So,

Now, conserving momentum along the north direction, we have:

Initial momentum = Final momentum

$m_1u_{1y}+m_2u_{2y}=(m_1+m_2)v_y\\\\1000\times 30+0=(1000+1250)v_y\\\\v_y=\frac{30000}{2250}\\\\v_y=\frac{40}{3}\ m/s$

There is no component of initial velocity for second car in north direction, as it is moving in the east direction. So, $u_{2y}=0$ .

The magnitude of final velocity is given as:

$|\vec{v}|=\sqrt{(v_x)^2+(v_y)^2}\\\\|\vec{v}|=\sqrt{(\frac{65}{3})^2+(\frac{40}{3})^2}\\\\|\vec{v}|=\sqrt{\frac{5825}{9}}=25.44\ m/s$

The direction is given as:

$\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{\frac{40}{3}}{\frac{65}{3}})=31.61^\circ$

So, the final velocity has a magnitude of 25.44 m/s and is at 31.61° north of east.

7 0

3 years ago

Suppose that the voltage across the resistor is held constant at 40 volts. If the resistance is steadily decreasing at a rate of

hjlf

Answer:

The current is increasing at a rate of 0.32 ampere per second.

Explanation:

The voltage of the resistor is modelled after Ohm's Law, which states that voltage is directly proportional to current:

$V = i\cdot R$ (1)

Where:

$V$ - Voltage, measured in volts.

$i$ - Current, measured in amperes.

$R$ - Resistance, measured in ohms.

An expression for the rate of change in voltage is found by Differential Calculus:

$\frac{dV}{dt} = \frac{di}{dt}\cdot R +i\cdot \frac{dR}{dt}$

$\frac{dV}{dt} = \frac{di}{dt}\cdot R + \frac{V}{R}\cdot \frac{dR}{dt}$ (2)

Where:

$\frac{dV}{dt}$ - Rate of change in voltage, measured in volts per second.

$\frac{di}{dt}$ - Rate of change in current, measured in amperes per second.

$\frac{dR}{dt}$ - Rate of change in resistance, measured in ohms per second.

If we know that $\frac{dV}{dt} = 0\,\frac{V}{s}$ , $R = 5\,\Omega$ , $V = 40\,\Omega$ and $\frac{dR}{dt} = -0.2\,\frac{\Omega}{s}$ , then the rate of change in current is:

$5\cdot \frac{di}{dt}-1.6 = 0$ (3)

$\frac{di}{dt} = 0.32\,\frac{A}{s}$

The current is increasing at a rate of 0.32 ampere per second.

5 0

3 years ago

Indicate whether each of the following statements is true or false. In a coal-fired power plant, chemical energy is first conver

denis23 [38]

Answer:

1. In a coal-fired power plant, chemical energy is first converted to thermal energy. TRUE.

The chemical energy in the coal is converted to thermal energy when the coal is burnt to produce steam.

2. In a coal-fired power plant, approximately 2/5 of original energy in the coal is lost to heat. FALSE.

Approximately 3/5 of the original energy is lost not 2/5 so this statement is false.

3. The molecules in cold air move faster than in hot air. FALSE.

Molecules with more heat move faster than molecules with less. Molecules in cold air therefore, will move slower than molecules in hot air.

6 0

3 years ago

A steel plate weighing 180 lb with center of gravity at point G is supported by a roller at point A, a bar DE, and a horizontal

melamori03 [73]

Answer:

Therefore, the force supported by the hydraulic cylinder is = -70.01 lb

Thus, the force supported by the bar = -233.84 lb

The reaction force supported by the reaction of the roller = 341.31 lb

Explanation:

The attached diagrams is shown in the file below:

From there; taking moments about point A

$\sum M__A }=0$

- 40 (180) - (80) $F_E$ Cos 37° + 55

-7200 + $F_E$ (-80 Cos 37° + 55 sin 37° ) = 0

= - 7200 - 30.79 $F_E$

- 30.79 $F_E$ = 7200

$F_E$ = $-\frac{7200}{30.79}$

$F_E$ = -233.84 lb

Thus, the force supported by the bar = -233.84 lb

Taking the equilibrium of forces in the vertical direction

$\sum f_y = 0$

$F_E$ sin 37 + $F_A$ cos 20 - $F_G$ = 0

- 233.84 sin 37 + $F_A$ cos 20 - 180 = 0

$F_A$ cos 20 = 320.73

$F_A$ = $\frac{320.73}{cos 20}$

$F_A$ = 341.31 lb

The reaction force supported by the reaction of the roller = 341.31 lb

Taking the equilibrium of forces on the horizontal direction.

$\sum f_y = 0$

$F_E$ cos 37 - $F_{CB} + F_A$ sin 20° = 0

-233.84 cos 37 - $F_{CB}$ + 341.31 sin 20° = 0

-186.75 - $F_{CB}$ + 116.74 = 0

- $F_{CB}$ -70.01 = 0

- $F_{CB}$ = 70.01

$F_{CB}$ = - 70.01 lb

Therefore, the force supported by the hydraulic cylinder is = -70.01 lb

7 0

4 years ago

How does sound travel?

Doss [256]

Answer:

Sound vibrations travel in a wave pattern, and we call these vibrations sound waves. Sound waves move by vibrating objects and these objects vibrate other surrounding objects, carrying the sound along. ... Sound can move through the air, water, or solids, as long as there are particles to bounce off of.

Explanation:

8 0

3 years ago

Read 2 more answers