Physics is explicitly involved in studying which of these activities?

You're driving down the highway late one night at 21 m/s when a deer steps onto the road 35 m in front of you. Your reaction tim

andreev551 [17]

Speed with which initially car is moving is 21 m/s

Reaction time = 0.50 s

distance traveled in the reaction time d = v t

d = 21 * 0.50 = 10.5 m

deceleration after this time = -10 m/s^2

now the distance traveled by the car after applying bakes

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

$0 - 21^2 = 2(-10)d$

$d = 22.05 m$

so total distance moved before it stop

d = 22.05 + 10.5 = 32.55 m

so the distance from deer is 35 - 32.55 = 2.45 m

now to find the maximum speed with we can move we will assume that we will just touch the deer when we stop

so our distance after brakes are applied is d = 35 - 10.5 = 24.5 m

again by kinematics

$v_f^2 - v_i^2 = 2 ad$

$0 - v^2 = 2(-10)(24.5)$

$v = 22.1 m/s$

so maximum speed would be 22.1 m/s

5 0

3 years ago

An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic

malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

$\omega_f = \omega_0 + \alpha t$

Where,

$\omega_f =$ Final Angular Velocity

$\omega_0 =$ Initial Angular velocity

$\alpha =$ Angular acceleration

t = time

The relation between the tangential acceleration is given as,

$a = \alpha r$

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

$\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s$

$\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s$

$t = 11s$

Replacing the previous equation with our values we have,

$\omega_f = \omega_0 + \alpha t$

$9.8436 = 6.5973 + \alpha (11)$

$\alpha = \frac{9.8436- 6.5973}{11}$

$\alpha = 0.295rad/s^2$

The tangential velocity then would be,

$a = \alpha r$

$a = (0.295)(0.2)$

$a = 0.059m/s^2$

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

$\omega_f^2=\omega_0^2+2\alpha\theta$

Replacing with our values and re-arrange to find $\theta,$

$\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}$

$\theta = \frac{9.8436^2-6.5973^2}{2*0.295}$

$\theta = 90.461rad$

That is equal in revolution to

$\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev$

The linear displacement of the system is,

$x = \theta*(2\pi*r)$

$x = 14.397*(2\pi*\frac{0.25}{2})$

$x = 11.3m$

5 0

3 years ago

What does the color of a star determine

Irina-Kira [14]

The color of light that a star emits is somewhat related to its age, whereas the color of light that we actually <em>observe</em> from a star is related to the speed at which it's moving with respect to us.

4 0

3 years ago

Jasmine makes a note card about series circuits. she records the points shown. a 2 column table with 1 row. the column on the le

Bess [88]

Jasmine need to correct the error by switching the headings on the columns adding the title parallel circuits.

<h3>What is a parallel circuit?</h3>

A parallel circuit is a circuit in which the components are connected to a common junction. This implies that if one bulb goes out in a parallel connection, all the bulbs will go out.

As such, Jasmine need to correct the error by switching the headings on the columns adding the title parallel circuits.

Learn more about parallel connection: brainly.com/question/12400458

#SPJ4

7 0

2 years ago

5. How much heat is generated when an

Mila [183]

Answer:

I HOPE THIS IS CORRECT

Explanation:

Power of water =2 kw=2000w

Mass of water =200kg

difference in temperature ΔT=70−10=60oC

Concept

energy required to heat the water = energy given by water in time t=pt

energy required to increase tempeature of water by 60oC,Q=msΔT

S= specific heat =4200J/kgoC

pt=msΔT

2000×t=200×4200×60

t=25200

or t=25.2×103sec.

6 0

3 years ago