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3 years ago

An object is held at rest and allowed to drop How far does the object fall in 3.3 seconds?

1 answer:

6 0

After x seconds, an object will fall $\frac{1}{2}at^2$ where a is acceleration due to gravity and t is time

so when t=3.3
the distance it will fall is $(\frac{1}{2})(9.8\frac{m}{s^2})(3.3m)^2$ =53.361m
it will fal 53.361 meters

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A stone is dropped into a river from a bridge 41.7 m above the water. Another stone is thrown vertically down 1.80 s after the f

hram777 [196]

Answer:

31.75 m/s

Explanation:

h = 41.7 m

Let the initial velocity of the second stone is u

Let the time taken to reach to the bottom by the first stone is t then the time taken by the second stone to reach the ground is t - 1.8.

For first stone:

Use second equation of motion

$h=ut+\frac{1}{2}gt^2$

Here, u = 0, g = 9.8 m/s^2 and t be the time and h = 41.7

So, 41.7= 0 + 0.5 x 9.8 x t^2

41.7 = 4.9 t^2

t = 2.92 s ..... (1)

For second stone:

Use second equation of motion

$h=ut+\frac{1}{2}gt^2$

Here, g = 9.8 m/s^2 and time taken is t - 1.8 = 2.92 - 1.8 = 1.12 s, h = 41.7 m and u be the initial velocity

$h=u\left ( t-1.8 \right )+4.9\left ( t-1.8 \right )^2$ .... (2)

By equation the equation (1) and (2), we get

$41.7=1.12 u +4.9 \times 1.12^{2}$

u = 31.75 m/s

5 0

3 years ago

A 1.5 kg spherical ball is has a radius of 50 cm is rotating with angular velocity of 12 revolutions per minute. Determine the r

kykrilka [37]

Answer:

K.E = 0.0075 J

Explanation:

Given data:

Mass of the ball = 1.5 kg

radius, r = 50 cm = 0.5 m

Angular speed, ω = 12 rev/min = (12/60) rev/sec = 0.2 rev/sec

Now,

the kinetic energy is given as:

K.E = $K.E=\frac{1}{2}I\omega^2$

where,

I is the moment of inertia = mr²

on substituting the values, we get

$K.E=\frac{1}{2}\times1.5\times0.5^2\times0.2^2$

K.E = 0.0075 J

3 0

3 years ago

What potential difference is required to cause 4.00 a to flow through a resistance of 330 ω?

Alisiya [41]

We can solve the problem by using Ohm's law, which states that an Ohmic conductor the following relationship holds:
$\Delta V = I R$
where
$\Delta V$ is the potential difference applied to the resistor
I is the current flowing through it
R is the resistance

In our problem, I=4.00 A and $R=330 \omega$ , so the potential difference is
$\Delta V = IR=(4.00 A)(330 \omega)=1320 V$

7 0

3 years ago

Since the moon has less mass than the earth what happens to objects on the moon

postnew [5]

When you drop an object on the moon, it falls to the ground.
But it only falls about 1/6 as fast as it falls on Earth.

4 0

3 years ago

what happens to a circuits resistance, voltage, and current when you increase the diameter of the wire in the circuit?

zhannawk [14.2K]

Answer:

Option D.

Resistance (R) decreases

Voltage (V) is constant

Current (I) increases

Explanation:

We'll begin by writing an equation relating resistance and diameter of a wire. This is given below:

R = ρL/A ......... (1)

A = πr² (since the wire is circular in shape)

r = d/2

A = πr² = π(d/2)²

A = πd²/4

Substitute the value of A into equation 1

R = ρL/A

R = ρL ÷ A

R = ρL ÷ πd²/4

R = ρL × 4/πd²

R = 4ρL /πd²

Where:

R is the resistance of the wire.

ρ is the resistivity of the wire.

L is the length of the wire.

A is the cross sectional area of the wire.

r is the radius.

d is the diameter of the wire

From equation (1) above, we can say that the resistance (R) is inversely proportional to the square of the diameter of the wire. This implies that an increase in the diameter of the wire will result in a decrease of the resistance. Also, a decrease in the diameter of the wire will result in an increase in the resistance of the wire.

1. Since the diameter of the wire is increase, therefore, the resistance of the wire will decrease.

2. From ohm's law,

V = IR

Divide both side by I

R = V/I

Where:

R is the resistance

V is the voltage

I is the current

From the above equation, the resistance (R) is directly proportional to the voltage (V) and inversely proportional to the current (I).

If we keep the voltage constant, this means that an increase in the resistance will lead to a decrease in the current. Also, a decrease in the resistance will lead to an increase in the current.

Since the resistance of the wire decrease, the current will increase.

From the illustrations made above, an increase in the diameter of the wire will lead to:

1. Decrease in resistance.

2. Voltage is constant.

3. Increase in current.

5 0

3 years ago