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MrRissso [65]
4 years ago
14

Which has greater kinetic energy, a car traveling at 30 km/HR or a car of half the mass traveling at 60 km/HR? a. The 60 km/HR c

arb. The 30 km/HR carc. Both have the same kinetic energyd. More information is needed about the distance traveled
Physics
1 answer:
artcher [175]4 years ago
6 0

Answer:

a.The 60 km/HR car

Explanation:

Kinetic Energy: This can be defined as the energy of a body due to motion. The S.I unit of kinetic energy is Joules (J).

It can be expressed mathematically as

Ek = 1/2mv²......................... Equation 1

Where Ek = kinetic energy, m = mass, v = velocity.

(i) A car travelling at 30 km/hr, with a mass of m,

Ek = 1/2(m)(30)²

Ek = 450m J.

(ii) A car travelling at 60 km/hr, with a mass of m/2

Ek = 1/2(m/2)(60)²

Ek = 900m J.

Thus , the car travelling at 60 km/hr at half mass has a greater kinetic energy to the car traveling at 30 km/hr at full mass.

The right option is a.The 60 km/HR car

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Three forces act on a moving object. One force has a magnitude of 83.7 N and is directed due north. Another has a magnitude of 5
LekaFEV [45]

Answer:

  • |\vec{F}_3| = 102.92 \ N
  • \theta = 57 \° 24 ' 48''

Explanation:

For an object to move with constant velocity, the acceleration of the object must be zero:

\vec{a} = \vec{0}.

As the net force equals acceleration multiplied by mass , this must mean:

\vec{F}_{net} = m \vec{a} = m * \vec{0} = \vec{0}.

So, the sum of the three forces must be zero:

\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{0},

this implies:

\vec{F}_3  = - \vec{F}_1 - \vec{F}_2.

To obtain this sum, its easier to work in Cartesian representation.

First we need to define an Frame of reference. Lets put the x axis unit vector \hat{i} pointing east,  with the y axis unit vector \hat{j} pointing south, so the positive angle is south of east. For this, we got for the first force:

\vec{F}_1 = 83.7 \ N \ (-\hat{j}),

as is pointing north, and for the second force:

\vec{F}_2 = 59.9 \ N \ (-\hat{i}),

as is pointing west.

Now, our third force will be:

\vec{F}_3  = - 83.7 \ N \ (-\hat{j}) - 59.9 \ N \ (-\hat{i})

\vec{F}_3  =  83.7 \ N \ \hat{j}  + 59.9 \ N \ \hat{i}

\vec{F}_3  =  (59.9 \ N , 83.7 \ N )

But, we need the magnitude and the direction.

To find the magnitude, we can use the Pythagorean theorem.

|\vec{R}| = \sqrt{R_x^2 + R_y^2}

|\vec{F}_3| = \sqrt{(59.9 \ N)^2 + (83.7 \ N)^2}

|\vec{F}_3| = 102.92 \ N

this is the magnitude.

To find the direction, we can use:

\theta = arctan(\frac{F_{3_y}}{F_{3_x}})

\theta = arctan(\frac{83.7 \ N }{ 59.9 \ N })

\theta = 57 \° 24 ' 48''

and this is the angle south of east.

7 0
3 years ago
How do I find the maximum height?
MakcuM [25]
To find the maximum height you have to calculate the initial velocity in the y axis..
to do that resolve the velocity vector in the y direction
which is equal to [email protected]
27*√3/2 = 23.4 m/s
thus using kinematics.
v^2 = u^2 + 2as
23.4*23.4 /19.6 = s
27.96 that's 28 m approximately
4 0
3 years ago
Which type of electromagnetic radiation is most likely to be ionizing
zlopas [31]

Answer:

Xrays. i hope that helps

Explanation:

4 0
3 years ago
Read 2 more answers
A 10.-kilogram object, starting from rest, slides down a frictionless incline with a constant acceleration of 2.0 m/sec2 for fou
aniked [119]

Answer:

C) 100N

Explanation:

Formula for calculating the Weight of an object is expressed as;

Weight = mass × acceleration due to gravity

Given

Mass of the object = 10kg

Acceleration due to gravity = 9.81m/s²

Substitute into the formula above

Weight = 10×9.81

Weight = 98.1N

Hence the approximate weight of the object is 100N

5 0
3 years ago
Step by step process on how to solve this problem
rusak2 [61]

<u>Step-1:</u>  Remember or look up the formula for the force of gravity between two objects.    F = G · m₁ · m₂ / R²

<u>Step-2:</u>  Remember or look up the value of G.<em> </em> G = 6.67 x 10⁻¹¹ m³·kg/s²

<u>Step-3: </u> Write the numbers you know into the formula.

(1.989 x 10²⁰ Newtons) =

(6.67 x 10⁻¹¹ mtr³·kg/s²) · (5.9742 x 10²⁴ kg) · (moon mass) / (3.84 x 10⁸ mtr)²

<u>Step-4:</u>  Sit back, relax, take your time, look this mess over, carefully, end-to-end, and decide how to solve it for (moon mass) .

<u>Step-5:</u> Divide each side by (6.67x10⁻¹¹mtr³·kg/s²)·(5.9742x10²⁴kg)/(3.84x10⁸mtr)²:

Moon mass =

(1.989x10²⁰ Newtons)·(3.84x10⁸ mtr)²/(6.67x10⁻¹¹ mtr³·kg/s²)·(5.9742x10²⁴ kg)

<u>Step-6: </u> Crunch the numbers.  Be careful to KEEP all the units as you go along. When you're done, the units of your answer will be the first instant indication if you made a mistake.  You're looking for the MASS of the moon.  If the answer doesn't have units of 'kg', then that'll be an immediate red flag, telling you that there's been a mistake somewhere.

Moon mass =

(1.989x10²⁰ Newtons)·(3.84x10⁸ mtr)²/(6.67x10⁻¹¹ mtr³·kg/s²)·(5.9742x10²⁴ kg)

Collect the numbers, and collect the units:

Moon mass = (1.989x10²⁰ · (3.84x10⁸)² / (6.67x10⁻¹¹ · 5.9742x10²⁴)

(kg-mtr/s² · mtr²)/(mtr³·kg/s² · kg)

Moon mass = (1.989 · 3.84² x 10³⁶) / (6.67 · 5.9742 x 10¹³)

(kg-mtr/s² · mtr²)/(mtr³·kg/s² · kg)

Moon mass = 0.736 x 10²³ kg

Moon mass = 7.36 x 10²² kg

<u>Step 7: </u> Look it up in a book or online.  See if you're anywhere close.

When I search "moon mass" on Floogle, the first hit says

" 7.348 x 10²² kg " .

yay !  I'm satisfied.

6 0
3 years ago
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