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3 years ago

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If you drop an object from a height of 1.4 m, it will hit the ground in 0.53 s. If you throw a baseball horizontally with an ini

tial speed of 35 m/s from the same height, how long will it take the ball to hit the ground?

1 answer:

insens350 [35]3 years ago

5 0

Answer:

The ball to hit the ground in 0.53 s.

Explanation:

Given that,

Height = 1.4 m

Time t = 0.53 s

Initial speed = 35 m/s

We need to calculate the time when the ball to hit the ground

Using equation of motion

$s_{y}=u_{y}t-\dfrac{1}{2}gt^2+h_{0}$

Where, s= vertical height

u= vertical velocity

t = time

h = height

Put the value in equation

$0=0-\dfrac{1}{2}\times9.8\times t^2+1.4$

$t^2=\dfrac{1.4}{4.9}$

$t=\sqrt{\dfrac{1.4}{4.9}}$

$t=0.53\ s$

Hence, The ball to hit the ground in 0.53 s.

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A cricket player catches the ball leaning towards to the ground,why?

Vadim26 [7]

Answer:

Explanation:

As it’s difficult to catch it from up.

Gravitational force will pull us when we jump.

If gravity was not there, he could catch the ball. But he will float in the sky after that.

That’s the answer

3 0

3 years ago

A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w

spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

directly proportional to its length i.e $R\propto l$
inversely proportional to its cross section area i.e $R\propto \frac{1}{A}$

Therefore

$R=\rho\frac{l}{A}$

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

$A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2$

$R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }$

$\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }$ ......(1)

Again for tungsten:

$R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }$

$\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }$ ........(2)

Given that $R_1=R_2$ and $l_1=l_2$

Dividing the equation (1) and (2)

$\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}$

$\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}$ [since $R_1=R_2$ and $l_1=l_2$ ]

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}$

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}$

$\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}$

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

8 0

3 years ago

two resistors of 20 ohm each are connected in a parallel with a battery of 10V. The total current passing through circuit is

Varvara68 [4.7K]

Answer:

1 Ampere.

Explanation:

From the question given above, the following data were obtained:

Resistor 1 (R₁) = 20 ohm

Resistor (R₂) = 20 ohm

Voltage (V) = 10 V

Current (I) =?

Next, we shall determine the equivalent resistance in the circuit. This can be obtained as follow:

Resistor 1 (R₁) = 20 ohm

Resistor (R₂) = 20 ohm

Equivalent Resistance (R) =?

Since the resistors are in parallel connection, the equivalent resistance can be obtained as follow:

R = (R₁ × R₂) / (R₁ + R₂)

R = (20 × 20) / (20 + 20)

R = 400 / 40

R = 10 ohm

Finally, we shall determine the total current in the circuit. This can be obtained as illustrated below:

Voltage (V) = 10 V

Equivalent Resistance (R) = 10 ohm

Current (I) =?

V = IR

10 = I × 10

Divide both side by 10

I = 10 / 10

I = 1 Ampere

Therefore, the total current in the circuit is 1 Ampere.

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