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Seven seconds after a brilliant flash of lightning, thunder shakes the house. How far was the lightning strike from the house? Seven seconds after a brilliant flash of lightning, thunder shakes the house. How far was the lightning strike from the house? Much farther away than two kilometers Much closer than one kilometer About two kilometers away About one kilometer away It is impossible to predict.

Answers

Answer 1

Answer:

Answer:

About two kilometers away

$\rm distance=2.401\ km$

Explanation:

Given:

The time gap between the light and sound to travel to the house, $t=7\ s$

Since the clouds are formed in the troposphere region of the atmosphere which extends from 8 kilometers to 12 kilometers above the earth-surface and the velocity of light is 300000 kilometers per second so it is visible almost instantly, hence we neglect the time taken by the light to travel to the house from the clouds.

∴Distance between the lightning-strike and the house:

$\rm distance=v* t$

we have the speed of sound as: $v=343\ m.s^(-1)$

So,

$\rm distance=343* 7$

$\rm distance=2401\ m$

$\rm distance=2.401\ km$

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What is the angular width of a person's thumb viewed at arm's length? Assume that the width of the thumb is 17.3 mm and that the distance between the eyes and the thumb is 71.9 cm. Use the small-angle approximation and then convert the answer to degrees.
A rock is thrown vertically upward from some height above the ground. It rises to some maximum height and falls back to the ground. What one of the following statements is true if air resistance is neglected? The acceleration of the rock is zero when it is at the highest point. The speed of the rock is negative while it falls toward the ground. As the rock rises, its acceleration vector points upward. At the highest point the velocity is zero, the acceleration is directed downward. The velocity and acceleration of the rock always point in the same direction.

The asteroid 234 Ida has a mass of about 4 × 1016 kg and an average radius of about 16 km. What is the acceleration due to gravity on 234 Ida? Assume that the asteroid is spherical; use G = 6.67 × 10–11 Nm2/kg2.A. 1 cm/s2
B. 2 cm/s2
C. 5 cm/s2
D. 6 cm/s2

Answers

The asteroid 234 Ida has a mass of about 4×1016 kg and an average radius of about 16 km. The acceleration due to gravity will be 1.04 cm/s². Hence, option A is correct.

What is the acceleration due to gravity?

The acceleration an object experiences as a result of gravitational force is known as acceleration due to gravity. M/s² is its SI unit. Its vector nature—which includes both magnitude and direction—makes it a quantity. The unit g stands for gravitational acceleration. At sea level, the standard value of g on the earth's surface is 9.8 m/s².

The formula for the acceleration due to gravity is g=GM/r².

According to the question, the given values are :

Mass, M = 4 × 1016 kg or

M = 4 × 10¹⁶.

Radius, r = 16 km or,

r = 16000 meter.

G = 6.67 × 10⁻¹¹ Nm²/kg²

g = (6.67 × 10⁻¹¹ ) (4 × 10¹⁶) / 16000²

g = 0.0104 m/s² or,

g = 1.04 cm/s².

Hence, the acceleration due to gravity will be 1.04 m/s²

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Answer:

1 cm/s²

Explanation:

I just took the quiz

A uniform piece of wire, 20 cm long, is bent in a right angle in the center to give it an L-shape. How far from the bend is the center of mass of the bent wire?

Answers

Answer:TL;DR: 3.535 cm

Explanation:

Xcm = ΣxMoments/ΣMasses = (10*0 + 10*5)/(10+10) = 50/20 = 2.5 cm

by symmetry,

Ycm = 2.5 cm

The distance D from the point Xcm,Ycm to the origin is D = √(2.5²+2.5²) = 3.535 cm

Final answer:

The center of mass of the bent wire is approximately 11.18 cm from the bend.

Explanation:

In order to find the center of mass of the bent wire, we need to divide it into two segments: the horizontal segment and the vertical segment. The length of each segment is half of the total length of the wire, which is 20 cm, so each segment is 10 cm long.

The center of mass of the horizontal segment is located exactly at its middle point, which is 5 cm from the corner. The center of mass of the vertical segment is also located at its middle point, which is 10 cm from the corner. Since the horizontal and vertical segments are orthogonal, the distance from the bend to the center of mass of the bent wire is the hypotenuse of a right triangle with legs of length 5 cm and 10 cm. Using the Pythagorean theorem, we can calculate the distance:

d = sqrt(5^2 + 10^2) = sqrt(25 + 100) = sqrt(125) = 11.18 cm

Therefore, the center of mass of the bent wire is approximately 11.18 cm from the bend.

Three small objects are arranged along a uniform rod of mass m and length L. one of mass m at the left end, one of mass m at thecenter, and one of mass 2m at the right end. How far to the left or right of the rod's center should you place a support so that the rod
with the attached objects will balance there?

Answers

Answer:

See answer below

Explanation:

Hi there,

To get started, recall the Center of Mass formula for two masses:

$x_c_m = (m_1x_1+m_2x_2)/(m_1+m_2)$ where m is mass and x is displacement from the center of the shape.

Since masses at the center of a geometric shape have a displacement (x) value of 0, as the mass is already of the center, and does not affect Xcm. So, we can disregard the central mass, hence we use the above formula for two masses.

We can arbitrarily define left to be a negative (-) displacement, and vice versa for right direction. We proceed with the formula: $x_c_m=((-L/2)m+(L/2)2m)/(m+2m) =((L/2)(-m+2m))/(3m) \n x_c_m=(L(m))/(6m) =(L)/(6)$

Since we defined left (-) and right (+), we notice the center of mass is (+) value. This makes sense, as there is slightly more mass on the right side. Hence, you should place a support 1/6 of the rod's length away from the rod's center.

Study well and persevere.

thanks,

Final answer:

To balance the rod with the attached objects, place a support at a distance of L/3 from the left end of the rod.

Explanation:

To balance the rod with the attached objects, you need to place a support at a distance of L/3 from the left end of the rod. This is because the center of gravity of the system should be directly above the support.

The center of gravity is given by the equation xcg = (m*0 + m*L/2 + 2m*L)/ (m + m + 2m). Solving this equation, we get xcg = 2L/3. Therefore, the support should be placed at a distance of L/3 from the left end of the rod.

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A proton is moving horizontally halfway between two parallel plates that are separated by 0.60 cm. The electric field due to the plates has magnitude 720,000 N/C between the plates away from the edges. If the plates are 5.6 cm long, find the minimum speed of the proton if it just misses the lower plate as it emerges from the field.

Answers

Answer:

v = 4,244,699 m/s = (4.245 × 10⁶) m/s

Explanation:

The electric force on the proton is given by

F = qE

where q = charge on the proton = (1.602 × 10⁻¹⁹) C

E = Electric field = 720,000 N/C

F = (1.602 × 10⁻¹⁹ × 720000)

F = (1.153 × 10⁻¹³) N

But this force will accelerate the proton in this magnetic field in a form of trajectory motion.

We can obtain the acceleration using Newton's first law of motion relation

F = ma

m = mass of a proton = (1.673 × 10⁻²⁷) kg

a = (F/m)

a = (1.153 × 10⁻¹³)/(1.673 × 10⁻²⁷)

a = 68,944,411,237,298 m/s²

a = (6.894 × 10¹³) m/s²

This acceleration directs the proton from the positive plate to the negative plate, covering a distance of y = 0.006 m (the distance between the plates)

Using Equations of motion, we can obtain the time taken for the proton to move from the rest at the positive plate to the negative one.

u = initial velocity of the proton = 0 m/s

y = vertical distance covered by the proton = 0.006 m

a = acceleration of the proton in this direction = (6.894 × 10¹³) m/s²

t = time taken for the proton to complete this distance = ?

y = ut + (1/2) at²

0.006 = 0 + [(1/2)×(6.894 × 10¹³)×t²]

0.006 = (3.447 × 10¹³) t²

t² = (0.006)/(3.447 × 10¹³)

t² = 1.741 × 10⁻¹⁶

t = (1.32 × 10⁻⁸) s

Then we can then calculate the minimum speed to navigate the entire length of the plates without hitting the plates.

v = ?

x = 0.056 n

t = (1.32 × 10⁻⁸)

v = (x/t)

v = (0.056)/(1.32 × 10⁻⁸)

v = 4,244,699 m/s = (4.245 × 10⁶) m/s

Hope this Helps!!!

Answer:

v = 9.09×10⁵m/s

Explanation:

Given

d = the distance between plates = 0.6cm = 0.006

E = Electric field strength = 720000N/C

m =mass of the proton = 1.67 ×10-²⁷ kg

The

Electric potential energy of the field is converted into the the kinetic energy of the proton.

qV = 1/2mv²

But V = Ed

So q(Ed) = 1/2mv²

v² = 2qEd/m

v = √(2qEd/m)

v = √(2×1.6×10-¹⁹×720000×0.006/1.67×10-²⁷)

v = 9.09×10⁵m/s

Which best describes a reference frame?

Answers

Answer:

C a position from which something is observed

om edu 2021

Explanation:

Answer: A system or frame of reference are those conventions used by an observer (usually standing at a point on the ground) to be able to measure the position and other physical magnitudes.

Find the current if 20C of charge pass a particular point in a circuit in 10 seconds.