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nikdorinn [45]
3 years ago
15

What is the strength of the electric field of a point charge of magnitude +4.8 × 10^-19 C at a distance of 4.0 × 10^-3 m? A. 3.6

× 10^-4 N/C B. -2.7 × 10^-4 N/C C. -3.6 × 10^-4 N/C D. 2.7 × 10^-4 N/C
Physics
1 answer:
soldi70 [24.7K]3 years ago
8 0
Answer is D -  from the equation of electric field for point charge
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A 2.0 kg book is lying on a 0.78 m -high table. You pick it up and place it on a bookshelf 2.1 m above the floor.Part ADuring th
zloy xaker [14]

Given data:

* The mass of the book is 2 kg.

* The initial height of the book is 0.78 m.

* The final height of the book is 2.1 m.

Solution:

(A). The work done by the gravity on the book is,

W=mg(h_f-h_i)

where m is the mass, g is the acceleration due to gravity, h_i is the initial height and h_f is the final height,

The work is done in moving the object in upward direction where as the gravitational force is acting in the down ward direction. Thus, the value g (acceleration due to gravity) is taken as negative in this case.

Substituting the known values,

\begin{gathered} W=2\times(-9.8)\times(2.1-0.78) \\ W=-19.6\times1.32 \\ W=-25.9\text{ J} \end{gathered}

Thus, the work done by the gravitational force is -25.9 J.

(B). The work done by the hand on the moving the book is,

\begin{gathered} W_1=-W \\ =25.9\text{ J} \end{gathered}

Thus, the work done by the hand on the book is 25.9 J.

7 0
11 months ago
Lenses can be found in a variety of objects which example does not have a lens?
Burka [1]
Tbh ! All of them have lens but I would go with B.) Smart Phone 

Hope This Helps !!!
4 0
3 years ago
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A spring with a rest length of 0.7 m has a spring constant of 70 N/m. It is stretched and now has a length of 2.5 m. What is the
pentagon [3]

Answer:

<em>113.4 J</em>

Explanation:

<u>Elastic Potential Energy</u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

\displaystyle PE = \frac{1}{2}k(\Delta x)^2

The spring has a natural length of 0.7 m and a spring constant of k=70 N/m. When the spring is stretched to a length of 2.5 m, the change of length is

Δx = 2.5 m - 0.7 m = 1.8 m

The energy stored in the spring is:

\displaystyle PE = \frac{1}{2}70(1.8)^2

PE = 113.4 J

7 0
3 years ago
Choose all that apply. Solids, liquids, and gases can be distinguished by their:
vodomira [7]
I believe its by there shape
8 0
3 years ago
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A tiger leaps horizontally out of a tree that is 6.00 m high. If he lands 2.00 m from the base of the tree, calculate his initia
Musya8 [376]

Answer:

The initial speed of the tiger is 1.80 m/s

Explanation:

Hi there!

The equation of the position vector of the tiger is the following:

r = (x0 + v0 · t, y0 + 1/2 · g · t²)

Where:

r = position vector at a time t.

x0 = initial horizontal position.

v0 = initial horizontal velocity.

t = time.

y0 = initial vertical position,

g = acceleration due to gravity.

Let´s place the origin of the frame of reference on the ground at the point where the tree is located so that the initial position vector will be:

r0 = (0.00, 6.00) m

We can use the equation of the vertical component of the position vector to obtain the time it takes the tiger to reach the ground.

y = y0 + 1/2 · g · t²

When the tiger reaches the ground, y = 0:

0 = 6.00 m - 1/2 · 9.81 m/s² · t²

2 · (-6.00 m) / -9.81 m/s² = t²

t = 1.11 s

We know that in 1.11 s the tiger travels 2.00 m in the horizontal direction. Then, using the equation of the horizontal component of the position vector we can find the initial speed:

x = x0 + v0 · t

At t = 1.11 s, x = 2.00 m

x0 = 0

2.00 m = v0 · 1.11 s

2.00 m / 1.11 s = v0

v0 = 1.80 m/s

The initial speed of the tiger is 1.80 m/s

4 0
3 years ago
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