A student designed an experiment to demonstrate the movement of Earth plates. The steps of the experiment are listed below.

Absolute zero (K=0 or -273.15°C) is what temperature on the Farenheit scale? a) 459.4°F b) -301.43 °F c) 233.05°F d) -40.15°F e

ANEK [815]

Answer:

e) None of these is true

Explanation:

Given that

Temperature = 0 K

We know that relationship between kelvin and Farenheit scale

$\dfrac{K-273}{100}=\dfrac{F-32}{180}$

Now by putting the values

$\dfrac{K-273}{100}=\dfrac{F-32}{180}$

$\dfrac{0-273}{100}=\dfrac{F-32}{180}$

So F= - 459.67°F

So we can say that 0 K is equal to - 459.67°F.

So the our option e is correct.

3 0

4 years ago

Two blocks of masses 6 kg and 5.5 kg are

sp2606 [1]

When you squish the spring, you put some energy into it, and after the cord
burns and they go boing in opposite directions, that energy that you stored
in the spring is what gives the blocks their kinetic energy.

But linear momentum still has to be conserved. It was zero while they were
tied together and nothing was moving, so it has to be zero after they both
take off.

Momentum = (mass) x (velocity)

After the launch, the 5.5-kg moves to the right at 6.8 m/s,
so its momentum is
(5.5 x 6.8) = 37.4 kg-m/s to the right.

In order for the total momentum to be zero, the other block has to
carry the same amount of momentum in the opposite direction.

M x V = (6 x speed) = 37.4 kg-m/s to the left.

Divide each side by 6 : Speed = 37.4 / 6 =<em> 6.2333... m/s left</em>

(That number is (6 and 7/30) m/s .)

5 0

3 years ago

A car travelling at 5.0 m/s starts to speed up. After 3.0 s its velocity has increased to 11 m/s. a) What is its acceleration? (

lions [1.4K]

initially, the car is traveling at 5.0 m/s.

so, we know acceleration for changing velocity is :

a = (v-v $_{o}$ )/t ..........(i)

where v is the final velocity

v $_{o}$ is the initial velocity

t is the time taken to change velocity

Now, as per the question :

initial velocity, v $_{o}$ =5.0 m/s

final velocity, v =11 m/s

time taken, t = 3 s

putting the values in equation (i),

a = ( 11-5 )/3

a = 2 m/s²

Therefore, a, after 3 s, is <em>2 m/s².</em>

4 0

1 year ago

Banked Curves: A 600-kg car is going around a banked curve with a radius of 110 m at a steady speed of 24.5 m/s. What is the app

Aleks04 [339]

The angle of baking from the calculation is obtained as 30°.

<h3>What is banking?</h3>

The term banking refers to a means of preventing vehicles from skidding off the road at curves.

We know that the banking angle is obtained from;

θ = tan-1(v^2/rg)

v = 24.5 m/s

r = 110 m

g = 9.8 m/s^2

θ = tan-1(25^2/9.8 * 110)

θ = tan-1(625 /1078)

θ = 30°

Learn more about the banking angle:brainly.com/question/26759099?r

#SPJ1

8 0

2 years ago

The legs of a weight lifter must ultimately support the weights he has lifted. A human tibia (shinbone) has a circular cross sec

Sergio039 [100]

Answer:

249003822308.05008 N

Explanation:

F = Force

$\sigma$ = Breaking stress of bone = 150 MPa

$d_2$ = Outer diameter = 3.6 cm

$d_1$ = Inner diameter = 2.3 cm

Area of the bone is assumed to be a hollow cylinder

$A=\dfrac{\pi}{4}(d_2^2-d_1^2)$

Stress is given by

$\sigma=\dfrac{F}{A}\\\Rightarrow F=\sigma A\\\Rightarrow F=\dfrac{150\times 10^6}{\dfrac{\pi}{4}(0.036^2-0.023^2)}\\\Rightarrow F=249003822308.05008\ N$

The maximum weight the person can lift without breaking his legs is 249003822308.05008 N

3 0

4 years ago