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scoundrel [369]
3 years ago
15

What is the answer to 17 ‐ 12× ‐ 18 ‐ 5×

Physics
1 answer:
Nastasia [14]3 years ago
7 0
The answer is -17x-1. You have to combine like terms.
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A test charge of -1.4 x 10-7 coulombs experiences a force of 5.4 x 10-1 newtons. Calculate the magnitude of the electric field c
VARVARA [1.3K]

Answer:

3.86×10⁶ Newton/coulombs

Explaination:

Applying,

E = F/q....................... Equation 1

Where E = Electric Field, F  = Force, q = charge.

From the question,

Given: F = 5.4×10⁻¹ N, q = -1.4×10⁻⁷ coulombs

Substitute these values into equation 1

E = 5.4×10⁻¹/ -1.4×10⁻⁷

E = -3.86×10⁶ Newtons/coulombs

Hence the magnitude of the electric field created by the

negative test charge is 3.86×10⁶ Newton/coulombs

5 0
2 years ago
When looking at a graph what are the first 3 things you should do?
Valentin [98]

Answer: Look where the points are.

Explanation:

4 0
2 years ago
Read 2 more answers
Does potential energy increase,kinetic energy decrease when a book is placed on a shelf
DerKrebs [107]
Yes potential increases while kinetic decreases
3 0
3 years ago
The magnitude of each force is 208 N the force on the right is applied at an angle 36° and the mass of the block is 17 kg the co
djyliett [7]

Answer:

<em>11.06m/s²</em>

Explanation:

According to Newtons second law of motion

\sm F_x = ma_x\\F_m - F_f = ma_x\\mgsin \theta - \mu R mgcos \theta = ma_x\\

Given

Mass m = 17kg

Fm = 208N

theta = 36 degrees

g = 9.8m/s²

a is the acceleration

Substitute

208 - 0.148(17)(9.8)cos 36 = 17a

208 - 24.6568cos36 = 17a

208 - 19.9478 = 17a

188.05 = 17a

a = 188.05/17

a = 11.06m/s²

<em>Hence the  the magnitude of the resulting acceleration is 11.06m/s²</em>

6 0
2 years ago
PLEASE HELPPPPP
Ganezh [65]

The only graph that accurately depict the given motion is graph D.

The given parameters;

  • initial position of the man = 0
  • direction of the man's first displacement = backward
  • time of first motion, t₁ = 6 seconds
  • velocity of this first displacement = v₁
  • time without any motion (<em>zero movement</em>) = 6 seconds
  • direction of the second displacement = forward
  • velocity of second displacement = 2v₁

Let the acceleration of the first displacement = a

Acceleration of the second displacement = 2a

From the given graphs we can eliminate every graph without initial decrease or motion towards the negative direction.

The only options with initial motion towards the negative direction are;

  • <em>graph B</em>, and
  • <em>graph D</em>.

The difference between graph B and D;

  • in graph B there is a uniform motion for 6 seconds
  • in graph D there is no motion for 6 seconds (<em>this is obvious as the line fall directly on top of the horizontal axis maintaining a value of zero for 6 seconds</em>).

Thus, the only graph that accurately depict the given motion is graph D.

Learn more here: brainly.com/question/21095906

5 0
2 years ago
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