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damaskus [11]
3 years ago
7

The high-speed winds around a tornado can drive projectiles into trees, building walls, and even metal traffic signs. In a labor

atory simulation, a standard wood toothpick was shot by pneumatic gun into an oak branch. The toothpick mass was 0.12 g, its speed before entering the branch was 221 m/s, and its penetration depth was 16 mm. If its speed was decreased at a uniform rate, what was the magnitude of the force of the branch on the toothpick?
Physics
1 answer:
Travka [436]3 years ago
3 0

Answer:

F  = 183.153 N

Explanation:

given,

mass of the toothpick = 0.12 g = 0.00012 kg  

initial velocity = 227 m/s    

final velocity = 0 m/s      

penetration depth = 16 mm = 0.016 m        

using the equation of motion        

v² - u² = 2 a s                            

0 - u² = 2 a s                                      

- 221² = 2 × a × 0.016      

a = 1526281.25 m/s²            

Force is equal to            

F = m a

  = 0.00012 × 1526281.25

F  = 183.153 N

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liraira [26]

Answer:

1.2 × 10^27 neutrons

Explanation:

If one neutron = 1.67 × 10^-27 kg

then in 2kg...the number of neutrons

; 2 ÷ 1.67 × 10^-27

There are.... 1.2 × 10^27 neutrons

4 0
2 years ago
I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block
borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

T = 2\pi \sqrt{\frac{m}{k} }

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\

Now, determine the amplitude of oscillation, A;

E = \frac{1}{2} kA^2

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A =  \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm

Therefore, the amplitude of the oscillation is 2.82 cm

8 0
2 years ago
An electron moving with a velocity v⃗ = 5.0 × 107 m/s i^ enters a region of space where perpendicular electric and a magnetic fi
stiv31 [10]

Answer:

Magnetic field, B=2\times 10^{-4}\ T

Explanation:

Given that,

Velocity of electron, v=5\times 10^7\ m/s

It enters  a region of space where perpendicular electric and a magnetic fields are present.

Magnitude of electric field, E=10^4\ V/m

We need to find the magnetic field will allow the electron to go through the region without being deflected.

Magnetic force on the electron, F_m=qvB\ sin\theta.......(1)

Electric force on the electron, F = q E........(2)

From equation (1) and (2) we get:

qvB\ sin\theta=qE

B=\dfrac{E}{v}

B=\dfrac{10^4\ V/m}{5\times 10^7\ m/s}

B = 0.0002 T

or

B=2\times 10^{-4}\ T

Hence, this is the required solution.

3 0
2 years ago
Hello please help i’ll give brainliest
Masteriza [31]

Answer:

D, the lithosphere. (CRUST AND UPPER MANTLE)

Explanation:

A tectonic plate (also called lithospheric plate) is a massive, irregularly shaped slab of solid rock, generally composed of both continental and oceanic lithosphere. Plate size can vary greatly, from a few hundred to thousands of kilometers across; the Pacific and Antarctic Plates are among the largest. Plate thickness also varies greatly, ranging from less than 15 km for young oceanic lithosphere to about 200 km or more for ancient continental lithosphere (for example, the interior parts of North and South America).

Information found on:

<u>https://pubs.usgs.gov/gip/dynamic/tectonic.html#:~:text=A%20tectonic%20plate%20(also%20called,both%20continental%20and%20oceanic%20lithosphere.&text=Continental%20crust%20is%20composed%20of,such%20as%20quartz%20and%20feldspar.</u>

3 0
2 years ago
Read 2 more answers
a stone attached to 1m long string is moving with the speed of 5ms in a circle find the centripetal acceleration of the stone​
Dafna1 [17]

Answer:

The centripetal acceleration of the stone is 5 m/s²

Explanation:

The length of the string to which the stone is attached, r = 1 m

The speed with which the string is rotated, v = 5 m/s

The centripetal acceleration, a_c, is given as follows;

a_c = \dfrac{v^2}{r}

Therefore, the centripetal acceleration of the stone found as follows;

a_c = \dfrac{(5 \ m/s)^2}{1 \ m} = 5 \ m/s^2

The centripetal acceleration of the stone, a_c = 5 m/s².

5 0
2 years ago
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