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Bad White [126]
3 years ago
6

I hope everyone is ok rn with this tornado . prayers if anyone hurt

Physics
2 answers:
skad [1K]3 years ago
4 0
<h2><em>THERE WAS A TORNADO????????????</em></h2><h2><em></em></h2><h2><em>SAY WHAT?!</em></h2>
Marina86 [1]3 years ago
3 0
I hope everyone that everyone that is in the tornado stays safe. My prayers go out to them! have a nice day.
You might be interested in
If the potential due to a point charge is 490 V at a distance of 10 m, what are the sign and magnitude of the charge?
uysha [10]

Answer:

+5.4×10⁻⁷ C

Explanation:

Electric potential: This can be defined as the work done in bringing a unit charge from infinity to that point against the action of the field. The S.I unit of potential is volt (V)

The formula for potential is

V = kq/r............................ Equation 1

Where V = electric potential, k = proportionality constant, q = charge, r = distance.

making q the subject of the equation,

q = Vr/k............................ Equation 2

Given: V = 490 V, r = 10 m,

Constant: k = 9×10⁹ Nm²/C²

Substitute into equation 2

q = 490(10)/(9×10⁹)

q = 5.4×10⁻⁷ C

q = +5.4×10⁻⁷ C

Hence the charge is +5.4×10⁻⁷ C

7 0
3 years ago
A spring scale hung from the ceiling stretches by 6.4 cm when a 2.0 kg mass is hung from it. The 2.0 kg mass is removed and repl
dolphi86 [110]
In the first case, the force acting on the spring is the weight of the mass:
F=mg=(2.0 kg)(9.81 m/s^2)=19.6N
This force causes a stretching of x=6.4 cm=0.064 m on the spring, so we can use these data to find the spring constant:
k= \frac{F}{x}= \frac{19.6 N}{0.064 m}=306.3 N/m

In the second case, the first mass is replaced with a second mass, whose weight is
F=mg=(2.5 kg)(9.81 m/s^2)=24.5 N
And since we know the spring constant, we can calculate the new elongation of the spring:
x= \frac{F}{k}= \frac{24.5 N}{306.3 N/m}=0.080 m=8.0 cm
5 0
3 years ago
To practice Problem-Solving Strategy 7.2 Problems Using Mechanical Energy II. The Great Sandini is a 60.0-kg circus performer wh
sp2606 [1]

Answer:

v = 15.45 m/s

Explanation:

As per mechanical energy conservation we can say that here since friction is present in the barrel so we will have

Work done by friction force = Loss in mechanical energy

so we will have

W_f = (U_i + K_i) - (U_f + K_f)

here we know that

W_f = F_f . d

W_f = 40 \times 4

W_f = 160 J

Initial compression in the spring is given as

F = kx

4400 = 1100 x

x = 4 m

now from above equation

W_f = (\frac{1}{2}kx^2 + 0) - (mgh + \frac{1}{2}mv^2)

160 = (\frac{1}{2}1100(4^2) + 0) - (60 \times 9.8\times 2.50 + \frac{1}{2}(60)v^2)

160 = 8800 - 1470 - 30 v^2

v = 15.45 m/s

3 0
3 years ago
I need help
kupik [55]
I like to just keep writing them over and over on a page also if you remember them just before the exam than as soon as you start write them on the front of your test so you don’t forget them.
Hope this helps.
7 0
3 years ago
Read 2 more answers
QUESTION 12
Y_Kistochka [10]

Answer:

The target was higher than the cannon

Explanation:

Given that a cannon ball is launched with initial speed 56 m/s, and hits a target at speed 35 m/s.

The initial velocity must be 56 m/s, while the final velocity will be 35 m/s.

When an object is thrown upward with an initial velocity, the object continues to move upwards with decreasing in magnitude of velocity till it reaches the maximum height where the final velocity will be equal to zero.

In this question, since the final velocity is less than the initial velocity, we can conclude that the target is higher.

Therefore, the target was higher than the cannon.

5 0
4 years ago
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