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Triss [41]
3 years ago
9

a custodian pulls a vacuum 13.5 m with a 33.9 N force at a 55.0 degree angle , against a 14.2 N friction force . find the work d

one by the friction force
Physics
1 answer:
Elenna [48]3 years ago
4 0

Answer:

a. -191.7 J

Explanation:

The work done by a force is given by W = Fdcosθ where F = force applied and d = distance moved and θ is the angle between the force and distance moved.

Here, F = frictional force = 14.2 N

d = distance = 13.5 m

and θ = 180° since the direction of the frictional force and distance are opposite.

So, W = Fdcosθ = 14.2 × 13.5cos180° = -191.7 J

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What is the amount of matter in a substance?​
professor190 [17]

Answer:

mass

Explanation:

Mass (M) is the measure of the amount of matter in an object. Mass is measured in grams (g). Mass is measured on a balance by comparing the object against other objects with known masses.

6 0
2 years ago
Read 2 more answers
When a 4.60-kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.30 cm. (a) If
Alex787 [66]

Answer:

a) y = 0.0075 m

b) W = 1.569 J

Explanation:

See attachment for the solution

6 0
3 years ago
How much work do you do on a 15 N book in lifting it straight up for a distance of<br> 0.40 meters?
astra-53 [7]

Answer:

Work done, W = 6 J

Explanation:

It is given that,

Force of gravity acting on the book, weight of the book is 15 N

We need to find the work done in lifting the book straight up for a distance of  0.4 meters.

The weight of the book is acting in downward direction and the book is lifted straight up, it means angle between them is 180 degrees. Work done is given by :

W=Fd\cos180\\\\W=15\times 0.4\times \cos180\\\\W=-6\ J

So, the magnitude of work done in lifting the book is 6 joules.

7 0
3 years ago
An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see
mash [69]

Answer:

(7.90 × 10⁻¹⁵) J

Explanation:

The electric force exerted on the elecrron by rhe electric field is given by

F = qE

where |q| = charge on the particle = (1.602 × 10⁻¹⁹) C

E = magnitude of the electric field = (2.9 × 10⁶) V/m or N/C

F = 1.602 × 10⁻¹⁹ × 2.9 × 10⁶ = (4.646 × 10⁻¹³) N

From Newton's first law of motion relation, we can obtain the acceleration this force confers on the electron

F = ma

m = mass of the electron = (9.11 × 10⁻³¹) kg

a = acceleration of the electron caused by the electric force = ?

(4.646 × 10⁻¹³) = (9.11 × 10⁻³¹) × a

a = (4.646 × 10⁻¹³)/(9.11 × 10⁻³¹)

a = (5.10 × 10¹⁷) m/s²

Now, using the equations of motion, we can obtain the velocity with which the electron reaches the positive plate

u = initial velocity of the electron = 0 m/s (since the electron was initially at rest)

v = final velocity of the electron = ?

a = acceleration of the electron = (5.10 × 10¹⁷) m/s²

y = distance covered by the electron = 1.7 cm = 0.017 m

v² = u² + 2ay

v² = 0² + 2(5.10 × 10¹⁷)(0.017)

v² = (1.734 × 10¹⁶)

v = 131,677,182.5 m/s = (1.32 × 10⁸) m/s

Kinetic energy with which the electron hits the positive plate = (1/2)(m)(v²) = (1/2)(9.11 × 10⁻³¹)(1.32 × 10⁸)² = (7.90 × 10⁻¹⁵) J

Hope this Helps!!!

3 0
3 years ago
Could I get help plz
Sloan [31]

Answer:

440hz

Explanation:

saxophone a plays at 430hz and a frequency of 5 beats per second can be heard so saxophone b is playing at a frequency 10hz louder than saxophone A making it 440hz

3 0
3 years ago
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