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3 years ago

Which dog has the most kinetic energy

1 answer:

4 0

The formula to find the kinetic energy is:

Ek= 1/2 × m × v^2

1. Ek= 1/2×15×3^2
= 67.5 J

2.Ek= 1/2×8×4^2
=64 J

3.Ek= 1/2×12×5^2
= 150 J

4.Ek= 1/2×10×6^2
= 180 J

So the fourth dog has the most kinetic energy.

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Nine steps for developing scientific principles are

xxTIMURxx [149]

Answer:

1)Observe a phenomenon

2)Ask a question/ start inferring

3)Form a hypothesis

4)Create an experiment

5)Collect data

6)Compare results

7)Analyze

8)Report findings

9)Compare with other experiments

7 0

2 years ago

The image shows mountains in Alaska.

zalisa [80]

Answer:

A syncline is visible.

Explanation:

A syncline is visible.

8 0

3 years ago

Read 2 more answers

Compare the maximum rate of heat transfer to the basal metabolic rate by converting a bmr of 88 kcal/hr into watts. what is the

elena-14-01-66 [18.8K]

Explanation :

It is given that,

BMR i.e basal metabolic rate is 88 kcal/hr. So, BMR in watts is converted by the following :

We know that, 1 kilocalorie = 4184 joules

So, $1\ kcal/h=\dfrac{1\times 4184\ J}{3600\ sec}$

$1\ kcal/h=1.16\ J/sec$

J/sec is nothing but watts.

So, $1\ kcal/h=1.16\ watts$

and $88\ kcal/h=88\times 1.16\ watts = 102.08\ watts$

So, it can be seen that the body can accommodate a modes amount of activity in hot weather but strenuous activity would increase the metabolic rate above the body's ability to remove heat.

8 0

3 years ago

A projectile is fired with an initial velocity of 120.0 meters per second at an angle, θ above the horizontal. If the projectile

k0ka [10]

Answer:

θ = 62.72°

Explanation:

The projectile describes a parabolic path:

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x = x₀+ vx*t Formula (1)

vx = v₀x

Where:

x: horizontal position in meters (m)

x₀: initial horizontal position in meters (m)

t : time (s)

vx: horizontal velocity in m/s

v₀x: Initial speed in x in m/s

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis are:

y= y₀+(v₀y)*t - (1/2)*g*t² Formula (2)

vfy= v₀y -gt Formula (3)

Where:

y: vertical position in meters (m)

y₀ : initial vertical position in meters (m)

t : time in seconds (s)

v₀y: initial vertical velocity in m/s

vfy: final vertical velocity in m/s

g: acceleration due to gravity in m/s²

Data

v₀ = 120 m/s , at an angle θ above the horizontal

v₀x= 55 m/s

x-y components of the initial velocity ( v₀)

v₀x = v₀*cosθ Equation (1)

v₀y = v₀*sinθ Equation (2)

Calculating of the angle θ

We replace data in the Equation (1)

55 = 120*cosθ

cosθ = 55 / 120

$\theta = cos^{-1}( \frac{55}{120} )$

θ = 62.72°

3 0

3 years ago

An inattentive driver is traveling 18.0 m/s when he notices a red light ahead. his car is capable of decelerating at a rate of 3

ryzh [129]

Speed of the car given initially

v = 18 m/s

deceleration of the car after applying brakes will be

a = 3.35 m/s^2

Reaction time of the driver = 0.200 s

Now when he see the red light distance covered by the till he start pressing the brakes

$d_1 = v* t$

$d_1 = 18* 0.200 = 3.6 m$

Now after applying brakes the distance covered by the car before it stops is given by kinematics equation

$v_f^2 - v_i^2 = 2 a s$

here

vi = 18 m/s

vf = 0

a = - 3.35

so now we will have

$0^2 - 18^2 = 2*(-3.35)(s)$

$s = 48.35 m$

So total distance after which car will stop is

$d = d_1 + d_2$

$d = 48.35 + 3.6 = 51.95 m$

So car will not stop before the intersection as it is at distance 20 m

3 0

3 years ago