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3 years ago

What is net force? A. The fundamental forces B. The unified force C. The gravitational force D. The vector sum of the forces

Physics

2 answers:

avanturin [10]3 years ago

8 0

It is the vector sum of the forces

dedylja [7]3 years ago

7 0

Net force is the vector sum of the forces.

Answer: Option D

Explanation:

The net force can be referred as the sum total of all the forces exerted up on a body. The word net refers to the total, hence net force refers to the total force which exerted upon the object.

A force is related as the displacement experienced by an object due to external factors or interaction acting up on the object. Option A,B and C are irrelevant as they are separate forces exerted up on the object.

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Little Tammy lines up to tackle Jackson to (unsuccessfully) prove the law of conservation of momentum. Tammy’s mass is 34.0 kg a

Naily [24]

Answer:

So Tammy must move with speed 4.76 m/s in opposite direction of Jackson

Explanation:

As per law of conservation of momentum we know that there is no external force on it

So here we can say that initial momentum of the system must be equal to the final momentum of the system

now we have

$m_1v_1 + m_2v_2 = 0$

final they both comes to rest so here we can say that final momentum must be zero

now we have

$34 v + 54 (3 m/s) = 0$

$v = -4.76 m/s$

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2 years ago

The learning about Universe is called ________ ...

ASHA 777 [7]

Answer:

The learning about Universe is called

Astronomy.

Explanation:

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2 years ago

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It took an American football lineman 14 seconds to complete 10 repetitions of a 102 kg bench press. If the bar had to travel 1.4

DIA [1.3K]

Answer:

1050.651 W

Explanation:

m = Mass of bench press = 102 kg

h = Height of lift = 1.47 m

t = Time taken = 14 seconds

g = Acceleration due to gravity = 9.81 m/s²

The work done is

$W=mgh\\\Rightarrow W=102\times 9.81\times 1.47\\\Rightarrow W=1470.9114\ J$

For 10 times

$W=10\times 1470.9114\\\Rightarrow W=14709.114\ J$

Power is given by

$P=\dfrac{W}{t}\\\Rightarrow P=\dfrac{14709.114}{14}\\\Rightarrow P=1050.651\ W$

The power is 1050.651 W

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3 years ago

Which moon phase comes before the phase in this picture?

hodyreva [135]

Waxing gibbous
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2 years ago

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You stand at the top of a deep well. To determine the depth, D, of the well you drop a rock from the top of the well and listen

Paladinen [302]

Answer:

(A)

$\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s \right )D+t_t^2v_s^2=0$

(B) D=54.71 m

Explanation:

Free Fall

When a particle is dropped in free air, it starts falling to the ground with an acceleration equal to the gravity. If one wanted to know the height of launching, it can indirectly be measured by the time it takes to reach the ground by the formula

$\displaystyle D=\frac{gt^2}{2}$

Solving for t

$\displaystyle t=\sqrt{\frac{2D}{g}}$

If we are taking into consideration the time we can hear the sound it makes when hitting the ground (or water in this case), we must also consider the speed of the sound for the time it takes to reach back our ears. That time can be computed from the basic equation for the speed

$\displaystyle t=\frac{D}{v_s}$

(A)

The total measured time is the sum of both times and it's given as $t_t=3.5\ seconds$

$\displaystyle t_t=\sqrt{\frac{2D}{g}}+\frac{D}{v_s}$

From this equation we'll manage to compute D

First, we isolate the square root

$\displaystyle \sqrt{\frac{2D}{g}}=t_t-\frac{D}{v_s}$

Let's square both sides

$\displaystyle \frac{2D}{g}=t_t^2-2t_t\frac{D}{v_s}+\frac{D^2}{v_s^2}$

Multiplying by $v_s^2$

$\displaystyle \frac{2Dv_s^2}{g}=t_t^2v_s^2-2t_tDv_s+D^2$

Rearranging and factoring

$\boxed{\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s\right )D+t_t^2v_s^2=0}$

Now, let's put in numbers:

$g=9.8\ m/s^2,\ v_s=345\ m/s,t_t=3.5\ sec$

$\displaystyle D^2-\left (\frac{2(345)^2}{9.8}+2(3.5)(345)\right )D+(12.25)345^2=0$

Computing all the coefficients:

$\displaystyle D^2-26,705.82D+1,458,056.25=0$

Solving for D, we have two possible solutions:

$D=54.71,\ D=26,651.11$

The second solution is called "extraneous", since it comes from squaring an equation, which can introduce non-valid (or external) solutions. It's impossible, given the conditions of the problem, that the well could be 26.5 km deep. So we'll keep the only solution as.

D=54.71 m

Let's prove our calculations by computing both times:

$\displaystyle t_1=\sqrt{\frac{2(54.71)}{9.8}}=3.34\ sec$

$\displaystyle t_2=\frac{54.71}{345}=0.16\ sec$

We can see their sum is 3.5 seconds, 3.34 of which were taken to reach the bottom of the well, and 0.16 sec took the sound to reach the top.

3 0

3 years ago