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2 years ago

At what temperature is Fahrenheit scale reading equal to twice of celsius?

1 answer:

5 0

At -24.613 F<span> temperature is the Fahrenheit scale reading equal to twice that of the Celsius and at </span>160 F<span> temperatuar is the Fahrenheit scale reading equal to half that of the Celsius? So, 160 degrees Celsius will equal 320 degrees Fahrenheit.</span>

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Which of the following would produce the most power?

Fantom [35]

Answer:

A mass of 10 kilograms lifted 10 meters in 5 seconds.

Explanation:

Power can be defined as the energy required to do work per unit time.

Mathematically, it is given by the formula;

$Power = \frac {Energy}{time}$

But Energy = mgh

Substituting into the equation, we have

$Power = \frac {mgh}{time}$

Given the following data;

Mass = 10kg

Height = 10m

Time = 5 seconds

We know that acceleration due to gravity is equal to 9.8 m/s²

$Power = \frac {10*9.8*10}{5} = 490 Watts$

Hence, a mass of 10 kilograms lifted 10 meters in 5 seconds would produce the most power.

6 0

2 years ago

a ballistic pendulum is used to measure the speed of high-speed projectiles. A 6 g bullet A is fired into a 1 kg wood block B su

Galina-37 [17]

Answer:

(a) v-bullet = 399.04 m/s

(b) I = 2.38 kg m/s

Explanation:

(a) To calculate the initial speed of the bullet, you first take into account that the kinetic energy of both wood block and bullet, just after the bullet impacts the block, is equal to the potential gravitational energy of block and bullet when the cord is at 60° respect to the vertical.

The potential energy is given by:

$U=(M+m)gh$ (1)

U: potential energy

M: mass of the wood block = 1 kg

m: mass of the bullet = 6g = 6.0*10^-3 kg

g: gravitational constant = 9.8m/s^2

h: distance to the ground

The distance to the ground is calculate d by using the information about the length of the cord and the degrees of the cord respect to the vertical:

$h=l-lsin\theta\\\\h=2.2m-2,2m\ sin60\°=0.29m$

The potential energy is:

$U=(1kg+6*10^{-3}kg)(9.8m/s^2)(0.29m)=2.85J$

Next, the potential energy is equal to kinetic energy of the block and the bullet at the beginning of its motion:

$U=\frac{1}{2}(M+m)v^2\\\\v=\sqrt{2\frac{U}{M+m}}=\sqrt{2\frac{2.85J}{1kg+6*10^{-3}kg}}=2.38\frac{m}{s}$

Next, you use the momentum conservation law, in order to calculate the speed of the bullet before the impact:

$Mv_1+mv_2=(M+m)v$ (2)

v1: initial velocity of the wood block = 0m/s

v2: initial speed of the bullet

v: speed of bullet and block = 2.38m/s

You solve the equation (2) for v2:

$M(0)+mv_2=(M+m)v$

$v_2=\frac{M+m}{m}v=\frac{1kg+6*10^{-3}kg}{6*10^{-3}kg}(2.38m/s)\\\\v_2=399.04\frac{m}{s}$

The speed of the bullet before the impact with the wood block is 399.04 m/s

(b) The impulse is gibe by the change in the velocity of the block, multiplied by the mass of the block:

$I=M\Delta v=M(v-v_1)=(1kg)(2.38m/s-0m/s)=2.38kg\frac{m}{s}$

The impulse is 2.38 kgm/s

(c) The force on the cord after the impact is equal to the centripetal force over the block and bullet. That is:

$T=F_c=(M+m)\frac{v^2}{l}=(1.006kg)\frac{(2.38m/s)^2}{2.2m}=2.59N$

The force on the cord after the impact is 2.59N

4 0

3 years ago

Consider 3.5 kg of austenite containing 0.95 wt% c and cooled to below 727°c (1341°f). (a) what is the proeutectoid phase? (b) h

vladimir2022 [97]

A. The proeutectoid phase is Fe₃c because 0.95 wt/c is greater than the eutectoid composition which is 0.76 wt/c

b. We determine how much total territe and cementite form, we apply the lever rule expressions yields.
Wx = (fe₃c-co/cfe₃ c-cx = 6.70- 0.95/6.70- 0.022 = 0.86
The total cementite
Wfe₃C = 10-Cx/ Cfe₃c -Cx = 0.95 - 0.022/6.70 - 0.022 = 0.14
The total cementite which is formed is
(0.14) × (3.5kg) = 0.49kg

c. We calculate the pearule and the procutectoid phase which cementite form the equation
Ci = 0.95 wt/c
Wp = 6.70 -ci/6.70 - 0.76 = 6.70 -0.95/6.70 - 0.76 = 0.97
0.97 corresponds to mass.
W fe₃ C¹ = Ci - 0.76/5.94 = 0.03
∴ It is equivalent to
(0.03) × (3.5) = 0.11kg of total of 3.5kg mass.

4 0

2 years ago

a car moving at 11 m/s crashes into an obstacle and stops in 0.26s. compute the Force that a seatbelt exerts on a 21-kg child to

Natalka [10]

Answer:

890 N

Explanation:

Acceleration is change in velocity over change in time.

a = Δv / Δt

a = (11 m/s − 0 m/s) / 0.26 s

a = 42.3 m/s²

Force is mass times acceleration.

F = ma

F = (21 kg) (42.3 m/s²)

F ≈ 890 N

4 0

3 years ago

I need help with this someone help me?

Brilliant_brown [7]

Graph A so answer B also why isn’t answer a with graph A and B with graph B etc like that’s just confusing lol

3 0

2 years ago