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belka [17]
3 years ago
15

Tangential velocity 2. Parabolic pathway 3. Projectile 4. Centripetal acceleration 5. Centripetal force a. acceleration towards

the center caused by the centripetal force b. a force which keeps a body moving with a uniform speed along a circular path and is directed along the radius towards the center c. a curved path followed by projectiles d. an object projected through space, traveling in two dimensions, that accelerates vertically due to gravity e. the instantaneous velocity of a body moving in a circular path
Physics
1 answer:
Stels [109]3 years ago
4 0

1. Tangential velocity:

<em>e) the instantaneous velocity of a body moving in a circular path.</em>

2. Parabolic pathway

<em>c. a curved path followed by projectiles</em>

3. Projectile

<em>d) an object projected through space, traveling in two dimensions, that accelerates vertically due to gravity.</em>

4. Centripetal acceleration

<em>a) acceleration towards the center caused by the centripetal force</em>

5. Centripetal force

<em>b) a force which keeps a body moving with a uniform speed along a circular path and is directed along the radius towards the center</em>

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Ghella [55]
Sound waves are longitudal waves meaning they go back and forth
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3 years ago
A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º. What is the coefficient of
Alex_Xolod [135]

Given :

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.

To Find :

The coefficient of static friction between the box and the plane.

Solution :

Vertical component of force :

mg\ sin\ \theta =  120\times 10 \times sin\ 47^\circ{}=877.62 \ N

Horizontal component of force(Normal reaction) :

mg\ cos\ \theta =  120\times 10 \times cos\ 47^\circ{}=818.40 \ N

Since, box is on the verge of slipping :

mg\ sin\ \theta= \mu(mg \ cos\ \theta)\\\\\mu = tan \ \theta\\\\\mu = tan\ 47^o\\\\\mu = 1.07

Therefore, the coefficient of static friction between the box and the plane is 1.07.

Hence, this is the required solution.

7 0
4 years ago
Four charges are on the four corners of a square. Q1 = +5μC, Q2 = -10μC, Q3 = +5μC, Q4 = -10μC. The side length of the square is
Marat540 [252]

Answer:

Explanation:

Electric field due to a point charge Q at a point at distance d is given by the relation

E = \frac{K\times Q}{d^2}

Since Q1 and Q2 are of the same magnitude and distance , so they will create eletric field of same magnitude. Similarly field due to rest of the charges will also be same.

The charges are situated on the corners of a square in such a way that

equal charges of Q1 and Q3 are situated on the diametrically  opposite corners of the square. Fields due to these two charges will be equal and opposite in direction. Therefore net field due to these two  charges will be zero.  

On the same ground, we can say that field due to Q2 and Q4 at the centre will be equal and opposite and therefore they will cancel out each other. Net field at the centre will be zero

Overall, net field due to all the four charges will be zero

3 0
4 years ago
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Levart [38]
Its Kinetic, hope this helps you
7 0
3 years ago
Is it possible for the momentum of a system consisting of two carts on a low-friction track to be zero even if both carts are mo
choli [55]

Answer:

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Explanation:

The total momentum of the system of two carts is given by:

p=m_1 v_1 + m_2 v_2

where

m1, m2 are the masses of the two carts

v1, v2 are the velocities of the two carts

Let's remind that v (the velocity) is a vector, so its sign depends on the direction in which the cart is moving.

We want to know if it is possible that the total momentum of the system can be zero, so it must be:

p=0\\m_1 v_1 + m_2 v_2 = 0\\m_1 v_1 = -m_2 v_2

From this equation, we see that this condition can only occur if v1 and v2 have opposite signs. Opposite signs mean opposite directions: therefore, the total momentum can be zero if the two carts are moving into opposite directions.

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3 years ago
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