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miss Akunina [59]
3 years ago
9

With your hand parallel to the floor and your palm upright, you raise a 3-kg book upward with an acceleration of 2 m/s2. what is

the magnitude of force that your hand exerts on the book while it is accelerating?
Physics
2 answers:
Marianna [84]3 years ago
8 0

Answer:

35 N

Explanation:

We are using Newton's law of motion.

Here is what we know:

(T-w)= ma

w = 3 kg (we will change to Newtons in a second)

a = 2m/s^2

rearrange equation to solve for T (tension)

T = ma+w

all we need is mass

Now to get the correct mass we need to divide its weight by the gravitational field strength(in our case it is just gravity) but first we want our weight of 3kg to be in Newtons.

1 kg  = 9.81 N

w = (3)(9.81) = 29.4N

Now that our weight is 29.4N we need to divide that by gravity.

m = \frac{29.4}{9.8}

Now that we have mass we can plug in our numbers

(\frac{29.4}{9.8}*2)+29.4 = 35.4N

Plugging that into our calculator gives us the answer 35.4 or 35 N

(feel free to correct me if my reasonings for any of this stuff is incorrect)

kirza4 [7]3 years ago
3 0

35N i don´t know why

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Lesechka [4]

Answer:

\Delta V = 0.053 A

Explanation:

Electric field in a given region is given by equation

E = \frac{A}{r^4}

as we know the relation between electric field and potential difference is given as

\Delta V = -\int E. dr

so here we have

\Delta V = - \int (\frac{A}{r^4}) .dr

\Delta V = \frac{A}{3r_1^3} - \frac{A}{3r_2^3}

here we know that

r_1 = 1.71 m  and r_2 = 2.89 m

so we will have

\Delta V = \frac{A}{3}(\frac{1}{1.71^3} - \frac{1}{2.89^3})

so we will have

\Delta V = 0.053 A

8 0
3 years ago
A circuit is set up such that it has a current of 8 A. What would be the new current if the resistance was increased by a factor
RUDIKE [14]

Answer:

4 A

Explanation:

The relationship between current, voltage and resistance in a circuit is given by Ohm's law:

V=RI

where

V is the voltage

R is the resistance

I is the current

The equation can also be rewritten as

I=\frac{V}{R}

from which we see that the current is inversely proportional to the resistance, R.

In this problem, the initial current is I = 8 A. Then the resistance is doubled:

R ' = 2R

So the new current is

I'=\frac{V}{R'}=\frac{V}{2R}=\frac{1}{2}(\frac{V}{R})=\frac{I}{2}=4 A

so the current is halved.

7 0
3 years ago
A 30.0 kg box hangs from a rope. What is the tension (in N) in the rope if the box has an initial velocity of 2.0 m/s and is slo
hjlf

Answer:

360 N

Explanation:

m = 30kg    u = 2 m/s     a = -2m/s/s

Since the object has an initial velocity of 2 m/s and acceleration of -2 m/s/s

the object will come to rest in 1 second but the force applied in that one second can be calculated by:

F = ma

F = 30 * -2

F = -60 N (the negative sign tells us that the force is acting downwards)

Now, calculating the force applied on the box due to gravity

letting g = -10m/s/s

F = ma

F = 30 * -10

F = -300 N (the negative sign tells us that the force is acting downwards)

Now, calculating the total downward force:

-300 + (-60) = -360 N

<em></em>

<em>Hence, a downward force of 360 N is being applied on the box and since the box did not disconnect from the rope, the rope applied the same amount of force in the opposite direction</em>

Therefore tension on the force = <u>360 N</u>

5 0
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kompoz [17]

Answer:

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Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

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Acceleration can simply be defined as the rate of change of velocity with time. Mathematically, it is expressed as:

a = (v – u) / t

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a is the acceleration.

v is the final velocity.

u is the initial velocity.

t is the time.

With the above formula, we can obtain the acceleration of the car as follow:

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Acceleration (a) =?

a = (v – u) / t

a = (10 – 0) / 20

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a = 0.5 m/s²

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When friction is too weak, like for instance when there is black ice- our center of gravity is displaced too quickly and we can fall. Likewise, if there is a lot of slush on the ground, cars can slip and slide.
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