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LUCKY_DIMON [66]
3 years ago
8

Which has more heat, a home oven at 500 degrees Celsius, or a one ton commercial oven at 500 degrees celsius?

Physics
1 answer:
alisha [4.7K]3 years ago
8 0
A commercial oven because it is more durable plus it’s more professional so they use more heat then we do at home
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Which of the following light waves will have the highest frequency?
ira [324]

Answer:

a. blue light

Explanation: hope this helps :)

8 0
3 years ago
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Two small frogs simultaneously leap straight up from a lily pad. Frog A leaps with an initial velocity of 0.551 m/s, while frog
goblinko [34]

Answer:

d = .076 m

Explanation:

The time for frog A can be calculated from  equation of motion

v_f = v_o + at

where v_f is final velocity, a is acceleration due to gravity

so from given data we have

-0.551= 0.551 + (-9.8)(t)

t = 0.112 sec

Now we will use that time for frog B

v_f = v_o + at

v_f = 1.23 + (-9.8)(0.112)

v_f = 0.128 m/s(Note its positive)

For the displacement

s = v_o t + 0.5at^2

s = (1.23)(0.112) + (.5)(-9.8)(0.112)^2

d = .076 m

6 0
3 years ago
A wave has a frequency of 2.0 Hz. if its traveling at a velocity of 20 m/s what it is the wavelength of the wave
kirill [66]
The correct answer is B 40m
8 0
3 years ago
What is the speed of a proton whose kinetic energy is 3.4 kev ?
Andreas93 [3]
The kinetic energy of the proton is 3.4 kev
1 kev = 1.602 × 10^-16 joules
therefore 3.4 kev is equivalent to;
3.4 ×  (1.602 ×10^-16)= 5.4468 × 10^-16 J
Kinetic energy is calculated by the formula 1/2mv² where m is the mass and v is the velocity.
Therefore V = √((2 × ( 5.4468×10^-16))/ (1.67 ×10^-27))
                    = 8.077 × 10^5 m/s

8 0
3 years ago
A 500 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s . The rocket engine, when it is fired, exer
ivanzaharov [21]

Answer:

The rocket has to be launched 8 m from the hoop

Explanation:

Let's analyze this problem, the rocket is on a car that moves horizontally, so the rocket also has the same speed as the car; The initial horizontal rocket speed is (v₀ₓ = 3.0 m/s).

On the other hand, when starting the engines we have a vertical force, which creates an acceleration in the vertical axis, let's use Newton's second law to find this vertical acceleration

    F -W = m a

    a = (F-mg) / m

    a = F/m  -g

    a = 7.0/0.500  - 9.8

    a = 4.2 m/s²

We see that we have a positive acceleration and that is what we are going to use in the parabolic motion equations

Let's look for the time it takes for the rocket to reach the height (y = 15m) of the hoop, when the rocket fires its initial vertical velocity is zero (I'm going = 0)

    y = v_{oy} t + ½ a t²

    y = 0 + ½ a t²

    t = √ 2y/a

    t = √( 2 15 / 4.2)

    t = 2.67 s

This time is also the one that takes in the horizontal movement, let's calculate how far it travels

    x = v₀ₓ t

    x = 3 2.67

    x = 8 m

The rocket has to be launched 8 m from the hoop

8 0
3 years ago
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