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zlopas [31]
3 years ago
6

Multiplying or dividing vectors by scalar results in?

Physics
1 answer:
Minchanka [31]3 years ago
6 0
You divide sort of the similar but vector is distance and displacement and scalar is just distance
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With good tires and breaks, a car traveling 50 mi/hr on dry pavement can travel 400 ft when the driver reacts to something he se
Bogdan [553]

Answer:6.72 m/s^2

Explanation:

Given

initial velocity u=50 mi/hr\approx 73.33 ft/s

Distance traveled before stopping s=400 ft

using equation of

v^2-u^2=2as

where v=final velocity

u=initial Velocity

a=acceleration

s=displacement

v=0 as car stops after travelling 400 ft

0-73.33^2=2\times a\times 400

a=-6.72 m/s^2

negative sign indicates it is deceleration

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3 years ago
A student performs an activity to study how electric current flows in a circuit. The student constructs two different circuits,
natta225 [31]

Native_Americans_in_the_United_State0000000000000000000000000

Explanation:

8 0
3 years ago
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In a heat engine if 1000 j of heat enters the system the piston does 500 j of work, what is the final internal energy of the sys
son4ous [18]

Answer:

2500 J

Explanation:

We can solve the problem by using the first law of thermodynamics:

\Delta U =U_f - U_i =Q-W

where

Uf is the final internal energy of the system

Ui is the initial internal energy

Q is the heat added to the system

W is the work done by the system

In this problem, we have:

Q = +1000 J (heat that enters the system)

W = +500 J (work done by the system)

Ui = 2000 J (initial internal energy)

Using these numbers, we can re-arrange the equation to calculate the final internal energy:

U_f = U_i + Q-W=2000 J+1000 J-500 J=2500 J

3 0
4 years ago
- Atoms of different elements:
TiliK225 [7]
(C) have the same heat capacity
3 0
3 years ago
A pickup truck is traveling down the highway at a steady speed of 30.1 m/s. The truck has a drag coefficient of 0.45 and a cross
Sav [38]

Answer:

The energy that the truck lose to air resistance per hour is 87.47MJ

Explanation:

To solve this exercise it is necessary to compile the concepts of kinetic energy because of the drag force given in aerodynamic bodies. According to the theory we know that the drag force is defined by

F_D=\frac{1}{2}\rhoC_dAV^2

Our values are:

V=30.1m/s

C_d=0.45

A=3.3m^2

\rho=1.2kg/m^3

Replacing,

F_D=\frac{1}{2}(1.2)(0.45)(3.3)(30.1)^2

F_D=807.25N

We need calculate now the energy lost through a time T, then,

W = F_D d

But we know that d is equal to

d=vt

Where

v is the velocity and t the time. However the time is given in seconds but for this problem we need the time in hours, so,

W=(807.25N)(30.1m/s)(3600s/1hr)

W=87.47*10^6J (per hour)

Therefore the energy that the truck lose to air resistance per hour is 87.47MJ

4 0
3 years ago
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