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lys-0071 [83]
4 years ago
12

Which of the following is located in the temperature climate zone?

Physics
1 answer:
barxatty [35]4 years ago
5 0
The correct answer is C. Taiga hope it helps ( :
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When using the max out method to determine muscular strength, you must first do a warm-up set and a complete set before attempti
dlinn [17]

Answer:

True

Explanation:

I took the quiz .

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3 years ago
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2. A Formula One race car speeds up from rest to 27.8 m/s in a distance of 25 meters.
Nadya [2.5K]

Answer:

Correct answer:   a = 15.46 m/s²

Explanation:

The formula for accelerated movement with the given data is:

V² - V₀² = 2 · a · d     where the initial velocity  V₀ and the final V  

Since the initial velocity V₀ is zero, the formula is:

V² = 2 · a · d  => a = V² / (2 · d) = 27.8² / (2 · 25) = 772.84 / 50 = 15.46 m/s²

a = 15.46 m/s²

God is with you!!!

4 0
3 years ago
In the year 2090, and Burt is still alive! Well, he’s really just a brain connected to a “life machine.” Since Burt has no body,
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The answer is b hard theory
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3 years ago
If the distance of a charged particle from a wire changes from 10cm to 20 cm , what happens to its magnetic field
postnew [5]

Answer:

i would say it decreases

Explanation:

bc as distance grows the fields decrease

IM NOT SURE tho

5 0
3 years ago
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A car of 1000 kg with good tires on a dry road can decelerate (slow down) at a steady rate of about 5.0 m/s2 when braking. If a
vredina [299]

(a) 4.0 s

The acceleration of the car is given by

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time interval

For this car, we have

v = 0 (the final speed is zero since the car comes to a stop)

u = 20 m/s is the initial velocity

a=-5.0 m/s^2 is the deceleration of the car

Solving the equation for t, we find the time needed to stop the car:

t=\frac{v-u}{a}=\frac{0-(20 m/s)}{-5.0 m/s^2}=4 s

(b) 40 m

The stopping distance of the car can be calculated by using the equation

v^2 - u^2 = 2ad

where

v = 0 is the final velocity

u = 20 m/s is the initial velocity

a = -5.0 m/s^2 is the acceleration of the car

d is the stopping distance

Solving the equation for d, we find

d=\frac{v^2-u^2}{2a}=\frac{0^2-(20 m/s)^2}{2(-5.0 m/s^2)}=40 m

(c) -5.0 m/s^2

The deceleration is given by the problem, and its value is -5.0 m/s^2.

(d) 5000 N

The net force applied on the car is given by

F=ma

where

m is the mass of the car

a is the magnitude of the acceleration

For this car, we have

m = 1000 kg is the mass

a=5.0 m/s^2 is the magnitude of the acceleration

Solving the formula, we find

F=(1000 kg)(5.0 m/s^2)=5000 N

(e) 2.0\cdot 10^5 J

The work done by the force applied by the car is

W=Fd

where

F is the force applied

d is the total distance covered

Here we have

F = 5000 N

d = 40 m (stopping distance)

So, the work done is

W=(5000 N)(40 m)=2.0\cdot 10^5 J

(f) The kinetic energy is converted into thermal energy

Explanation:

when the breaks are applied, the wheels stop rotating. The car slows down, as a result of the frictional forces between the brakes and the tires and between the tires and the road. Due to the presence of these frictional forces, the kinetic energy is converted into thermal energy/heat, until the kinetic energy of the car becomes zero (this occurs when the car comes to a stop, when v = 0).

8 0
4 years ago
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