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2 years ago

What quantity is represented by a unit called a newton (N)?

Physics

1 answer:

zavuch27 [327]2 years ago

5 0

$\textsf {C. Force}$

$\textsf {The quantity which is represented by a unit called Newton (N)}\\\textsf {is Force.}$

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What is the lift (in newtons) due to Bernoulli's principle on a wing of area 76 m2 if the air passes over the top and bottom sur

AveGali [126]

Answer:

So lift will be 30.19632 N

Explanation:

We have given area of the wing $a=76m^2$

We know that density of air $d=1.29kg/m^3$

Speed at top surface $v_2=290m/sec$ and speed at bottom surface $v_1=150m/sec$

According to Bernoulli's principle force is given by

$F=A\times d\times \frac{v_2^2-v_1^2}{2}=76\times 1.29\times \frac{290^2-150^2}{2}=3019632N$

4 0

3 years ago

PLS ANSWER ASAP!!

Molodets [167]

b) between poles M1 and M2

Explanation:

From the expression, we can deduce that r is the distance between two magnetic poles M1 and M2.

The law of attraction between two magnetic poles states that:

<em> the force of attraction or repulsion between two magnetic poles is a function of the product of the strength of the magnetic poles and the square of the distance between the pole</em>s

Mathematically:

FM = K $\frac{M1 M2}{r^{2} }$

here r is the distance between the poles

FM is the magnetic force between the poles

M1 is the strength of the first magnetic pole

M2 is the strength of the second pole

K is the magnetic field constant

learn more:

magnetic pole brainly.com/question/2191993

#learnwithBrainly

8 0

3 years ago

The type of tectonic plate boundary that occurs when two plates are colliding into each other is known as a

ruslelena [56]

The answer is convergent boundary

6 0

3 years ago

Read 2 more answers

a ball of mass 100g moving at a velocity of 100m/s collides with another ball of mass 400g moving at 50m/s in same direction, if

klio [65]

Answer:

Velocity of the two balls after collision: $60\; \rm m \cdot s^{-1}$ .

$100\; \rm J$ of kinetic energy would be lost.

Explanation:

<h3>Velocity</h3>

Because the question asked about energy, convert all units to standard units to keep the calculation simple:

Mass of the first ball: $100\; \rm g = 0.1\; \rm kg$ .
Mass of the second ball: $400\; \rm g = 0.4 \; \rm kg$ .

The two balls stick to each other after the collision. In other words, this collision is a perfectly inelastic collision. Kinetic energy will not be conserved. The velocity of the two balls after the collision can only be found using the conservation of momentum.

Assume that the system of the two balls is isolated. Thus, the sum of the momentum of the two balls will stay the same before and after the collision.

The momentum of an object of mass $m$ and velocity $v$ is: $p = m \cdot v$ .

Momentum of the two balls before collision:

First ball: $p = m \cdot v = 0.1\; \rm kg \times 100\; \rm m \cdot s^{-1} = 10\; \rm kg \cdot m \cdot s^{-1}$ .
Second ball: $p = m \cdot v = 0.4\; \rm kg \times 50\; \rm m \cdot s^{-1} = 20\; \rm kg \cdot m \cdot s^{-1}$ .
Sum: $10\; \rm kg \cdot m \cdot s^{-1} + 20 \; \rm kg \cdot m \cdot s^{-1} = 30 \; \rm kg \cdot m \cdot s^{-1}$ given that the two balls are moving in the same direction.

Based on the assumptions, the sum of the momentum of the two balls after collision should also be $30\; \rm kg \cdot m \cdot s^{-1}$ . The mass of the two balls, combined, is $0.1\; \rm kg + 0.4\; \rm kg = 0.5\; \rm kg$ . Let the velocity of the two balls after the collision $v\; \rm m \cdot s^{-1}$ . (There's only one velocity because the collision had sticked the two balls to each other.)

Momentum after the collision from $p = m \cdot v$ : $(0.5\, v)\; \rm kg \cdot m \cdot s^{-1$ .
Momentum after the collision from the conservation of momentum: $30\; \rm kg \cdot m \cdot s^{-1}$ .

These two values are supposed to describe the same quantity: the sum of the momentum of the two balls after the collision. They should be equal to each other. That gives the equation about $v$ :

$0.5\, v = 30$ .

$v = 60$ .

In other words, the velocity of the two balls right after the collision should be $60\; \rm m \cdot s^{-1}$ .

<h3>Kinetic Energy</h3>

The kinetic energy of an object of mass $m$ and velocity $v$ is $\displaystyle \frac{1}{2}\, m \cdot v^{2}$ .

Kinetic energy before the collision:

First ball: $\displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.1\; \rm kg \times \left(100\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J$ .
Second ball: $\displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.4\; \rm kg \times \left(50\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J$ .
Sum: $500\; \rm J + 500\; \rm J = 1000\; \rm J$ .

The two balls stick to each other after the collision. Therefore, consider them as a single object when calculating the sum of their kinetic energies.