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iragen [17]
2 years ago
9

Sari made pancake batter. She has 1 cup of batter left. How much batter did she use?

Physics
1 answer:
Naily [24]2 years ago
5 0

She used 1 cup less than the amount that she made when she decided to make pancakes.  We have no idea how much she had to begin with.

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A skater of mass 40 kg is carrying a box of mass 5 kg. The skater has a speed of 5 m/s with respect to the floor and is gliding
Yuki888 [10]

For the part a) we need only the momentum of the box and we have the data to find it.

Momentum is given by,

p=mv

where clearly, p is the momentum, m the mass of the box and v is the velocity.

Substituting,

p=(5kg)(5m/s)\\p=25kg.m/s

For part b) we need an analysis of the situation. We understand that the box on a surface that has no friction will continue to rotate at the same speed previously defined. The box can only stop with friction, so,

p=(5kg)(5m/s)\\p=25kg.m/s

<em>It is the same that part a)</em>

8 0
2 years ago
The power of a machine is 6000 W. This machine is scheduled for design improvements. Engineers have reduced the
Georgia [21]

Explanation:

Work = power × time

W = (6000 W) (7.5 s)

W = 45,000 J

7 0
3 years ago
A single slit 1.4 mmmm wide is illuminated by 460-nmnm light. Part A What is the width of the central maximum (in cmcm ) in the
mamaluj [8]

Answer:

<h2> 1.643*10⁻⁴cm</h2>

Explanation:

In a single slit experiment, the distance on a screen from the centre point is expressed as y = \frac{\delta m \lambda d}{a} where;

\delta m is the first two diffraction minima = 1

\lambda is light wavelength

d is the distance of diffraction pattern from the screen

a is the width of the slit

Given \lambda = 460-nm = 460*10⁻⁹m

d = 5.0mm = 5*10⁻³m

a = 1.4mm = 1.4*10⁻³m

Substituting this values into the formula above to get width of the central maximum y;

y = 1*460*10⁻⁹ * 5*10⁻³/1.4*10⁻³

y = 2300*10⁻¹²/1.4*10⁻³

y = 1642.86*10⁻⁹

y = 1.643*10⁻⁶m

Converting the final value to cm,

since 100cm = 1m

x = 1.643*10⁻⁶m

x = 1.643*10⁻⁶ * 100

x = 1.643*10⁻⁴cm

Hence, the width of the central maximum in the diffraction pattern on a screen 5.0 mm away is  1.643*10⁻⁴cm

3 0
3 years ago
Play around with the moving man simulation. Then without using the simulation, sketch a predicted position
beks73 [17]
No idea man good luck though
3 0
3 years ago
WILL REWARD 20 more pts once solved
Mandarinka [93]

Explanation:

step 1. a diverging lens is "concave" on both side and always has a negative focal length

step 2. so 1/f = 1/s + 1/s' where f is the focal length, s is the object location, and s' is the image location (f, s, s' are all on the left side of the lens)

step 3. 1/-15 = 1/s + 1/-9 (image is virtual (negative))

step 4. 3/-45 = 1/s + 5/-45

step 5. s = 22.5cm (object is 22.5cm from lens)

step 6. s'/s = 9/22.5 ÷ 0.4 (magnification)

step 7. if the object is 4.5cm then the image is 4.5(0.4) = 1.8cm tall.

8 0
3 years ago
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