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2 years ago

Sunitha can type 1800 words in half an hour. What is her typing speed in words per minute?

Physics

1 answer:

Andre45 [30]2 years ago

6 0

Answer:

60words/minute

Explanation:

If Sunitha can type 1800 words in half an hour, this can be expressed as;

1800 words = 30 minutes

To get her typing speed per minute, we will use the formula

Speed = Number of words/Time used

Typing speed = 1800/30

Typing speed = 60words/minute

Hence her typing speed in words per minute is 60words/minute

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A cube of side 6.50 cm is placed in a uniform field E = 7.50 × 10^3 N/C with edges parallel to the field lines (field enters the

Svetllana [295]

Answer:

a) $\Phi_{net} = 0\,\frac{N\cdot m^{2}}{C}$ , b) $\Phi_{right} = -31.688\,\frac{N\cdot m^{2}}{C}$ , c) $\Phi_{left} = 31.688\,\frac{N\cdot m^{2}}{C}$

Explanation:

a) The net flux through the cube is:

$\Phi_{net}=-(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}+(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}$

$\Phi_{net} = 0\,\frac{N\cdot m^{2}}{C}$

b) The flux through the right face is:

$\Phi_{right}=-(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}$

$\Phi_{right} = -31.688\,\frac{N\cdot m^{2}}{C}$

c) The flux through the left face is:

$\Phi_{left}=(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}$

$\Phi_{left} = 31.688\,\frac{N\cdot m^{2}}{C}$

5 0

3 years ago

Read 2 more answers

5kg mass is attached to a point by a 35 meter rope. The mass isreleased and falls in a circle. If the only forces acting on the

mr_godi [17]

Answer:

Tension at the bottom of the swing T = 52.794N

Explanation:

Detailed explanation and calculation is shown in the image below

8 0

2 years ago

Choose two forces and compare and contrast these forces. These must be different forces than used in the prior question. Provide

marin [14]

Answer:

Explanation:

There are 4 forces. These are 1) Gravity, 2) Weak Nuclear Force, 3) Electromagnetism, and 4) Strong Nuclear Force.

Order of strength from weakest to strongest: Gravity, Weak Nuclear Force, Electromagnetism, Strong Nuclear Force

Type of Range:

Gravity - Unlimited range

Weak Nuclear Force - Limited range

Electromagnetism - Infinite range

Strong Nuclear Force - Limited Range

Found in:

Gravity - Exists between all objects with mass

Weak Nuclear Force - Governs over beta decays like the emission of electron or positron

Electromagnetism - the attraction found between particles that are electrically charged

Strong Nuclear Force - Found in atoms and subatomic particles. It is responsible for holding the atoms' nucleus together.

6 0

2 years ago

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22

alexgriva [62]

1) The law of motion of the projectile is
$h(t) = 2+22.5 t-4.9 t^2$
To find the velocity, we should compute the derivative of h(t):
$v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t$
So now we can calculate the speed at t=2 s and t=4 s:
$v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s$
$v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s$
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
$v(t)=0$
So we have
$22.5-9.8 t=0$
And solving this we find
$t=2.30 s$

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
$h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m$

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
$2+22.5t -4.9 t^2 = 0$
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
$t=4.68 s$
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
$v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36 m/s$
with negative sign, because it is directed downward.

8 0

3 years ago

What is the average acceleration of a tennis ball that has an initial velocity of 6.0 m/s [E] and a final velocity of 7.3 m/s [W

Marizza181 [45]

Given :

The average acceleration of a tennis ball that has an initial velocity of 6.0 m/s.

and a final velocity of 7.3 m/s.

It is in contact with a tennis racket for 0.094 s

To Find :

The average acceleration of the tennis ball.

Solution :

We know, average acceleration is given by :

$a_{avg}=\dfrac{Final \ velocity-Initial\ velocity}{Time\ Taken}\\\\a_{avg}=\dfrac{7.3-6.0}{0.094}\ m/s^2\\\\a_{avg}=13.83\ m/s^2$

Therefore, average velocity is given by 13.83 m/s².

Hence, this is the required solution.

7 0

2 years ago