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3 years ago

15

A 4.33 kg ball is dropped from the roof of a building 157.2 m high. While the ball is falling to Earth, a horizontal wind exerts

a constant force of 11.5 N on the ball. How long does it take to hit the ground? The acceleration of gravity is 9.81 m/s 2 . Answer in units of s.

1 answer:

laiz [17]3 years ago

3 0

Answer:

The ball takes 5.66 seconds to hit the ground.

Explanation:

In this problem, the horizontal wind that interferes with the ball doesn't affect the time in which the balls hit the ground.

To find this time this is the equation for kinetic energy:

$d=\frac{1}{2}gt^2$

Where d is the distance, g is the acceleration of gravity and t is the time that you have to find, you can solve this equation to find t:

$t=\sqrt{\frac{2d}{g}}$

Replace the values:

$t=\sqrt{\frac{2(157.2m)}{9.81/s}}$

t=5.66 s

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A 30 ohm resistor and a 20 ohm resistor are

Reptile [31]

The current that would pass through the 30 ohms resistor is 2 A.

<h3>What is electric current?</h3>

Electric current is the rate of flow of electric charge round a conductor.

To calculate the electric current that would pass through the 30 ohms resistor, we use the formula below

Formula:

I = V/Rt........... Equation 1

Where:

I = Electric current passing through the 30 ohms resistor
V = Voltage
Rt = Total or effective resistance of the resistors.

From the question,

Given:

V = 100 volts
Rt = (30+20) ohms (since both resistors are connected in series)

Rt = 50 ohms

Substitute these values into equation 1

I = 100/50
I = 2 A

Hence, The current that would pass through the 30 ohms resistor is 2 A.

Learn more about electric current here: brainly.com/question/1100341

8 0

3 years ago

A car drives around a curve with radius 539 m at a speed of 32.0 m/s. The road is banked at 5.00°. The mass of the car is 1.40 ×

HACTEHA [7]

Answer:

$f_r = 150.47 N$

Explanation:

given,

r = 539 m

v = 32 m/s

road banked at = 5°

∑ F_x

$\dfrac{mv^2}{r}= N sin \theta + f_r cos \theta$

∑ F_y = 0

$0 = N cos \theta - f_r sin \theta - mg$

$N = \dfrac{f_rsin \theta + mg}{cos \theta}$

$\dfrac{mv^2}{r}= (\dfrac{f_rsin \theta + mg}{cos \theta})sin \theta + f_r cos \theta$

= $f_r sin \theta tan \theta + mg tan \theta + f_r cos \theta$

$f_r = \dfrac{\dfrac{mv^2}{r}- mg tan\theta}{sin\theta tan \theta + cos \theta}$

$f_r = \dfrac{\dfrac{1.4\times 10^3 \times 32^2}{539}- 1.4\times 10^{3}\times 9.8 \times 0.087}{0.087 \times 0.087 + 0.996}$

$f_r = 150.47 N$

8 0

4 years ago

What is the current in a 160V circuit if the resistance is 20Ω

Mkey [24]

Answer:

I =V/R

I= 160/20 = 8 Amperes

3 0

3 years ago

PLEASE ANSWER, I NEED HELP

Scorpion4ik [409]

1) The gravitational force between Ellen and the moon is $1.56\cdot 10^{-3} N$

2) The two forces are equal, while the acceleration of the bus is smaller than the acceleration of the bicycle.

Explanation:

1)

The magnitude of the gravitational force between two objects is given by

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

$m_1, m_2$ are the masses of the two objects

r is the separation between them

In this problem, we have:

$m_1 = 47 kg$ is the mass of Ellen

$m_2 = 7.35\cdot 10^{22} kg$ is the mass of the moon

$r=3.84\cdot 10^8 m$ is the distance between Ellen and the moon

Substituting, we find the gravitational force between Ellen and the moon:

$F=(6.67\cdot 10^{-11})\frac{(47)(7.35\cdot 10^{22})}{(3.84\cdot 10^8)^2}=1.56\cdot 10^{-3} N$

2)

We can analyze the forces acting in the collision between the bus and the bicycle by using Newton's third law of motion, which states that:

"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"

Applied to our problem, this means that the force exerted by the bus on the bicycle during the collision (action force) is equal (and opposite) to the force exerted by the bicycle on the bus (reaction force).

Now let's analyze the accelerations of the two vehicles. We can find the acceleration of each vehicle by using Newton's second law:

$a=\frac{F}{m}$

where

a is the acceleration

F is the force exerted on the vehicle

m is the mass of the vehicle

As we said previously, the force F exerted on each of the two vehicles: so, the acceleration only depends on the mass. In particular, the acceleration is inversely proportional to the mass: therefore, the larger the mass of the vehicle, the smaller the acceleration. This means that the acceleration of the bus is smaller than the acceleration of the bicycle.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

And about Newton's third law:

brainly.com/question/11411375

#LearnwithBrainly

6 0

3 years ago

39. A dog runs on a waxed floor at an initial speed of 2 m/s. It slides to a stop with an

Sedbober [7]

Answer:

Explanation:

Use the one-dimensional equation

$v_f=v_0+at$ where vf is the final velocity of the dog, v0 is the initial velocity of the dog, a is the acceleration of the dog, and t is the time it takesto reach that final velocity. For us:

0 = 2 + -.43t and

-2 = -.43t so

t = 4.7 seconds

5 0

3 years ago