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nalin [4]
3 years ago
14

estion: Why is it important to use vector quantities and not just scalar quantities to describe the motion of an object? Vector

quantities deal with motion?​
Physics
1 answer:
Vera_Pavlovna [14]3 years ago
8 0

Answer:

Vector quantities are important in the study of motion. Some examples of vector quantities include force, velocity, acceleration, displacement, and momentum. The difference between a scalar and vector is that a vector quantity has a direction and a magnitude, while a scalar has only a magnitude. Vector, in physics, a quantity that has both magnitude and direction. It is typically represented by an arrow whose direction is the same as that of the quantity and whose length is proportional to the quantity's magnitude. A quantity which does not depend on direction is called a scalar quantity. Vector quantities have two characteristics, a magnitude and a direction. The resulting motion of the aircraft in terms of displacement, velocity, and acceleration are also vector quantities. A vector quantity is different to a scalar quantity because a quantity that has magnitude but no particular direction is described as scalar. A quantity that has magnitude and acts in a particular direction is described as vector.

Explanation:

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The value of specific heat for copper is 390 J/kg⋅°C, for aluminun is 900 J/kg⋅°C, and for water is 4186 J/kg⋅°C. What will be t
Irina-Kira [14]

Answer:

the equilibrium temperature Te = 19.9°C

Explanation:

Given;

specific heat for copper Cc is 390 J/kg⋅°C

for aluminun Ca is 900 J/kg⋅°C,

for water Cw is 4186 J/kg⋅°C

Mass of copper Mc= 265 g = 0.265kg

Temperature of copper Tc = 235°C

Mass of aluminium Ma = 135g = 0.135 kg

Temperature of aluminium Ta = 14.0°C

Mass of water Mw= 865 g = 0.865kg

Temperature of water Tw = 14.0°C

The equilibrium temperature can be derived by;

Te = (MaCaTa + McCcTc + MwCwTw)/(MaCa + McCc+ MwTw)

Substituting the values;

Te = ( 0.135×900×14 + 0.265×390×235 + 0.865×4186×14)/(0.135×900 + 0.265×390 + 0.865×4186)

Te = 19.939°C

Te = 19.9°C

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A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp.
Genrish500 [490]

<u>Answer:</u>

a) Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) Minimum speed must he drive off the horizontal ramp with 7° above the horizontal  = 23.93 m/s

<u>Explanation:</u>

a) The height of ramp = 1.5 meter

   Horizontal distance he must clear = 22 meter

   The car is having horizontal motion and vertical motion. In case of vertical motion the acceleration on the car is acceleration due to gravity.

   We have equation of motion, s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

 In case of vertical motion initial velocity = 0 m/s, acceleration = 9.8 m/s^2, we need to calculate time when displacement = 1.5 meter.

 1.5=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 0.553 seconds

So the car has to cover a distance of 22 meter in 2.119 seconds.

 So minimum speed required = 22/0.553 = 39.78 m/s

 Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) When the take of angle is 7⁰ the vertical speed of car is not zero = V sin 7 = 0.122 V

 So the in case of vertical motion we have initial velocity = 0.122 V, S = -1.5 meter( below ramp), acceleration = -9.8 m/s^2

Substituting

     -1.5=0.122V*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-0.122Vt-1.5=0

In case of horizontal motion

    Horizontal speed of car = V cos 7 = 0.993V

    So it has to travel 22 meter in t seconds

            0.993Vt = 22, Vt = 22.155 m

    Substituting in the equation 4.9t^2-0.122Vt-1.5=0

    We will get 4.9t^2-0.122*22.155-1.5=0\\ \\ t = 0.926 seconds

   Speed required = 22.155/0.926 = 23.93 m/s

  Minimum speed must he drive off the horizontal ramp with 7° above the horizontal  = 23.93 m/s

7 0
4 years ago
The temperature of a lead fishing weight rises from 26∘c to 38∘c as it absorbs 11.3 j of heat. what is the mass of the fishing w
Nimfa-mama [501]
Given:
ΔT = 38 - 26 = 12°C, temperature change
Q = 11.3 J, heat input
c = 0.128 J/(g-°C), specific heat of lead

Let m =  the mass of the lead.
Then
Q = m*c*ΔT
(m g)*(0.128 J/(g-°C))*(12 °C) = 11.3 J
1.536m = 11.3
m = 7.357 g

Answer:  7.36 g  (2 sig. figs)
8 0
3 years ago
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