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3 years ago

A stone dropped from the height of 90 m with velocity does it reach to the ground

Physics

1 answer:

Anon25 [30]3 years ago

7 0

Explanation and answer:

To solve this problem, we use the kinematics equation:

v^2-u^2=2aS .........................(1)

where

assuming the positive direction is downwards)

v = final velocity (when it reaches ground)

u = 0 = initial velocity (when it was dropped, or released, with zero vertical velocity).

a = 9.81 m/s^2 = acceleration due to gravity (on earth's surface)

S = 90m = distance travelled.

Substitute values into (1)

v^2 - 0^2 = 2*(9.81)*90

v^2 = 1765.8 m^2/s^2

v = sqrt(1765.8) m/s

= 42.02 m/s

= 42 m/s (to two significant figures).

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(a) What is the cost of heating a hot tub containing 1440 kg of water from 10.0°C to 40.0°C, assuming 75.0% efficiency to take h

BaLLatris [955]

Answer:

a) $E = 6.024\,USD$ , For m kilograms, it is 4184m J., 3600000 joules, b) $i = 88.200\,A$

Explanation:

a) The amount of heat needed to warm water is given by the following expression:

$Q_{needed} = m_{w}\cdot c_{w}\cdot (T_{f}-T_{i})$

Where:

$m_{w}$ - Mass of water, measured in kilograms.

$c_{w}$ - Specific heat of water, measured in $\frac{J}{kg\cdot ^{\circ}C}$ .

$T_{f}$ , $T_{i}$ - Initial and final temperatures, measured in $^{\circ}C$ .

Then,

$Q_{needed} = (1440\,kg)\cdot \left(4184\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (40^{\circ}C - 10^{\circ}C)$

$Q_{needed} = 180748800\,J$

The energy needed in kilowatt-hours is:

$Q_{needed} = 180748800\,J\times \left(\frac{1}{3600000}\,\frac{kWh}{J} \right)$

$Q_{needed} = 50.208\,kWh$

The electric energy required to heat up the water is:

$E = \frac{50.208\,kWh}{0.75}$

$E = 66.944\,kWh$

Lastly, the cost of heating a hot tub is: (USD - US dollars)

$E = (66.944\,kWh)\cdot \left(0.09\,\frac{USD}{kWh} \right)$

$E = 6.024\,USD$

The heat needed to raise the temperature a degree of a kilogram of water is 4184 J. For m kilograms, it is 4184m J. Besides, a kilowatt-hour is equal to 3600000 joules.

b) The current required for the electric heater is:

$i = \frac{Q_{needed}}{\eta \cdot \Delta V \cdot \Delta t}$

$i = \frac{180748800\,J}{0.75\cdot (220\,V)\cdot (3.45\,h)\cdot \left(3600\,\frac{s}{h} \right)}$

$i = 88.200\,A$

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3 years ago

A single-turn current loop carrying a 4.00 A current, is in the shape of a right-angle triangle with sides of 50.0 cm, 120 cm, a

pantera1 [17]

Given that,

Current = 4 A

Sides of triangle = 50.0 cm, 120 cm and 130 cm

Magnetic field = 75.0 mT

Distance = 130 cm

We need to calculate the angle α

Using cosine law

$120^2=130^2+50^2-2\times130\times50\cos\alpha$

$\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}$

$\alpha=\cos^{-1}(0.3846)$

$\alpha=67.38^{\circ}$

We need to calculate the angle β

Using cosine law

$50^2=130^2+120^2-2\times130\times120\cos\beta$

$\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}$

$\beta=\cos^{-1}(0.923)$

$\beta=22.63^{\circ}$

We need to calculate the force on 130 cm side

Using formula of force

$F_{130}=ILB\sin\theta$

$F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0$

$F_{130}=0$

We need to calculate the force on 120 cm side

Using formula of force

$F_{120}=ILB\sin\beta$

$F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63$

$F_{120}=0.1385\ N$

The direction of force is out of page.

We need to calculate the force on 50 cm side

Using formula of force

$F_{50}=ILB\sin\alpha$

$F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38$

$F_{50}=0.1385\ N$

The direction of force is into page.

Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.

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3 years ago

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Define malleability?<br>

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Answer:

Explanation:

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3 years ago

A 1.4-kg cart is attached to a horizontal spring for which the spring constant is 50 N/m . The system is set in motion when the

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Answer:

0.395 m

2.4 m/s

Explanation:

$m$ = mass of the cart = 1.4 kg

$k$ = spring constant of the spring = 50 Nm⁻¹

$x$ = initial position of spring from equilibrium position = 0.21 m

$v_{i}$ = initial speed of the cart = 2.0 ms⁻¹

$A$ = amplitude of the oscillation = ?

Using conservation of energy

Final spring energy = initial kinetic energy + initial spring energy

$(0.5) kA^{2} = (0.5) m v_{i}^{2} + (0.5) k x_{i}^{2} \\kA^{2} = m v_{i}^{2} + k x_{i}^{2} \\(50) A^{2} = (1.4) (2.0)^{2} + (50) (0.21)^{2} \\A = 0.395 m$

$m$ = mass of the cart = 1.4 kg

$k$ = spring constant of the spring = 50 Nm⁻¹

$A$ = amplitude of the oscillation = 0.395 m

$v_{o}$ = maximum speed at the equilibrium position

Using conservation of energy

Kinetic energy at equilibrium position = maximum spring potential energy at extreme stretch of the spring

$(0.5) m v_{o}^{2} = (0.5) kA^{2}\\m v_{o}^{2} = kA^{2}\\(1.4) v_{o}^{2} = (50) (0.395)^{2}\\v_{o} = 2.4 ms^{-1}$

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3 years ago

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