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3 years ago

A meteoroid that reaches earths surface without burning up is a____

Physics

1 answer:

Ad libitum [116K]3 years ago

3 0

<span>A meteoroid that reaches earths surface without burning up is a meteorite.</span>

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A segment is divided into two parts having lengths in the ratio of 5:3. If the difference between the length of the parts is 6",

Triss [41]

Answer:

15"

Explanation:

Let x be the length of the longer part, then the length of the shorter part is 3x/5, or also x - 6

therefore we have the following equation:

3x/5 = x - 6

We can multiply both sides by 5

3x = 5x - 6*5

2x = 30

x = 30/2 = 15"

So the length of the longer part is 15"

5 0

3 years ago

The ability of sculptural material to resist forces of pressure, like gravity, is called its __________.

Vitek1552 [10]

I think the answer would be tensile, I’m sorry if it’s wrong

3 0

3 years ago

A viscous fluid is flowing through two horizontal pipes. The pressure difference P1 - P2 between the ends of each pipe is the sa

777dan777 [17]

Answer:half of shorter Pipe

Explanation:

Fluid is Flowing through two horizontal pipes with pressure difference

$P_1-P_2=\Delta P$

Both pipes have same radius

Length of one Pipe is twice of other

Let Longer Pipe be denote by 1 and smaller by 2

From Hagen Poiseuille equation

$\Delta P=\frac{128\mu L\cdot Q}{\pi D^4}$

Where $\mu =$ viscosity of medium

L=length of Pipe

Q=discharge

D=diameter

For longer Pipe

$\Delta P=\frac{128\mu 2L\cdot Q_1}{\pi \cdot D^4}$ ----1

For smaller Pipe

$\Delta P=\frac{128\mu L\cdot Q_2}{\pi \cdot D^4}$ ------2

From 1 & 2 we get

$2L\cdot Q_1=L\cdot Q_2$

$Q_2=2Q_1$

volume flow rate of longer pipe is half of smaller pipe

6 0

3 years ago

A horizontal 745 N merry-go-round of radius

Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is 544J

Explanation:

Given :

Weight w = 745 N

Radius r = 1.45 m

Force = 56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62 = ?

Solution:

Step 1: Finding the Mass of merry-go-round

$m = \frac{ weight}{g}$

$m = \frac{745}{9.81 }$

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I = $0.5 \times m \times r^2$

Substituting the values

Moment of Inertia of solid cylinder I

=> $0.5 \times 76.02 \times (1.45)^2$

=> $0.5 \times 76.02\times 2.1025$

=> $79.91 kg.m^2$

Step 3: Finding the Torque applied T

Torque applied T = $F \times r$

Substituting the values

T = $56.3 \times 1.45$

T = 81.635 N.m

Step 4: Finding the Angular acceleration

Angular acceleration , $\alpha = \frac{Torque}{Inertia}$

Substituting the values,

$\alpha = \frac{81.635}{79.91}$

$\alpha = 1.021 rad/s^2$

Step 4: Finding the Final angular velocity

Final angular velocity , $\omega = \alpha \times t$

Substituting the values,

$\omega = 1.021 \times 3.62$

$\omega = 3.69 rad/s$

Now KE (100% rotational) after 3.62s is:

KE = $0.5 \times I \times \omega^2$

KE = $0.5 \times 79.91 \times 3.69^2$

KE = 544J

6 0

4 years ago

What force will a proton experience in a uniform electric field whose strength is 2.00 x 10^5 Newtons per Coulomb? *

Alenkasestr [34]

Answer: F =3.218 * 10^-11 N

Explanation: The relationship between electric field intensity and force is given by

F = Eq.

F= Force on electron

E = strength of electric field = 2*10^5

q= magnitude of proton charge = 1.609 * 10^-19c

F = 2* 10^5 * 1 609 * 10^-16

F = 3.218 * 10^-11 N

7 0

3 years ago