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natita [175]
3 years ago
7

Which two quantities are measured in the same units? (5 points)

Physics
2 answers:
Alex787 [66]3 years ago
7 0
C. Impulse and Momentum
andriy [413]3 years ago
4 0

Answer: Option (c) is the correct answer.

Explanation:

Impulse is the product of force applied on a body and the time period during which force has been exerted.

SI unit of impulse is newton second.

Whereas momentum is the product of mass and velocity of an object. Its SI unit is kg m/s.

Now, the units of both impulse and momentum are dimensionally equivalent to each other as follows.

      Impulse unit                  =              Momentum unit

    Newton second              =           \frac{kg \times m}{s}

Newton equals kg m/s^{2}.

\frac{kg \times m}{s \times s} \times s             =                     \frac{kg \times m}{s}

Cancelling out s from numerator and denominator of impulse unit. Therefore, we get the following.

  \frac{kg \times m}{s}                  =                   \frac{kg \times m}{s}

Thus, we can conclude that impulse and momentum are the two quantities which are measured in the same units.

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If we are viewing the atom in such a way that the electron's orbit is in the plane of the paper with the electron moving clockwi
Anastaziya [24]

Answer:

The magnitude  of the electric field is 5.1 \times 10^{11}  \frac{N}{C}

Explanation:

Given:

Charge of electron q = 1.6 \times 10^{-19}C

Separation between two charges r = 5.3 \times 10^{-11} m

For finding the magnitude of the electric field,

   E= \frac{kq}{r^{2} }

Where k = 9 \times 10^{9}

   E = \frac{9 \times 10^{9} \times 1.6 \times 10^{-19} }{(5.3 \times 10^{-11} )^{2} }

   E = 5.1 \times 10^{11} \frac{N}{C}

Therefore, the magnitude  of the electric field is 5.1 \times 10^{11}  \frac{N}{C}

5 0
4 years ago
The diagram shows the parabolic path of a projectile that leaves the foot of a kicker with a horizontal velocity of 15 m/s and a
rusak2 [61]

Answer:

I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.

Explanation:

In order to increase the horizontal distance covered by the ball, we need to examine the variables involved in the formula of range of projectile. The formula for the range of projectile is given as follows:

R = V₀² Sin 2θ/g

where, g is a constant on earth (acceleration due to gravity) and θ is the angle of ball with ground at the time of launching. The value of θ should be 45° for maximum range. In this case we do not know the angle so, we can not tell if we should change it or not.

The only parameter here which we can increase to increase the range is launch velocity (V₀). The formula for V₀ in terms of horizontal and vertical components is as follows:

V₀ = √(V₀ₓ² + V₀y²)

where,

V₀ₓ = Horizontal Velocity

V₀y = Vertical Velocity

Hence, it is clear from the formula that we can increase both the horizontal and vertical velocity to increase the initial speed which in turn increases the horizontal distance covered by the ball.

<u>Therefore, I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.</u>

4 0
3 years ago
Un observador se halla a 510m. de una pared. entre el observador y la pared, y a igual distancia de ambos, se realiza un disparo
ch4aika [34]

Answer:

a)    t = 0.75 s,  b)  t = 2.25 s

Explanation:

The speed of sound is constant in a material medium

           v = 340 m / s

we can use the relations of uniform motion to find the time

          v = x / t

          t = x / v

In the exercise, the observer's distance to the wall is indicated d = 510 m, it also indicates that the shot is fired at the midpoint

          x = d / 2

       

a) direct sound the distance from the observer to the screen is

           x = 510/2 = 255 m

           t = 255/340

           t = 0.75 s

b) echo sound.

In this case the sound reaches the wall bounces, the distance is

             x = d / 2 + d

             x = 3/2 d

             x = 3/2 510

             x = 765 m

the time is

              t = x / v

              t = 765/340

              t = 2.25 s

5 0
3 years ago
A stationary 500 kg tank fires a 20 kg miegile at 100 m/s. What is the velocity of the tank after the missile is fired? Assume t
dedylja [7]

Answer:

v₁ = 4 [m/s].

Explanation:

This problem can be solved by using the principle of conservation of linear momentum. Where momentum is preserved before and after the missile is fired.

P=m*v

where:

P = linear momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

(m_{1}*v_{1})=(m_{2}*v_{2})

where:

m₁ = mass of the tank = 500 [kg]

v₁ = velocity of the tank after firing the missile [m/s]

m₂ = mass of the missile = 20 [kg]

v₂ = velocity of the missile after firing = 100 [m/s]

(500*v_{1})=(20*100)\\v_{1}=2000/500\\v_{1}=4[m/s]

8 0
3 years ago
Because acceleration is a quantity that has both magnitude and direction, it is a(n)___________
swat32

Answer:

Vector Quantity

Explanation:

A Vector quantity is a quantity has both magnitude and direction while scalar quantity has only magnitude.

4 0
4 years ago
Read 2 more answers
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