Ask question

Ask question

All categories

Delicious77 [7]

3 years ago

6

A boy pulls a sled 46 m across a flat field with a force of 120 N. The force acts on the sled through a rope at a 23° angle with

the horizontal. What is the work the boy does on the sled? (Round your answer to two significant figures.)

1 answer:

ivann1987 [24]3 years ago

8 0

Answer:

5520

Explanation:

w=f×d

w=120×46

w=5.52×10^3

You might be interested in

during a workout, a football player pushes a blocking dummy a distance of 30 m. while pushing the dummy the same distance a seco

babymother [125]

$Power=\frac{Work}{Time}=\frac{Force\times distance}{time}$

If he wants to increase power, force must increase and decrease time.

8 0

4 years ago

Read 2 more answers

Predict the energy transformation that occurs when carson sands a piece of wood with sandpaper.

tigry1 [53]

Answer:

Explanation:

The hand provides Kinetic Energy in moving.

The KE is transformed to Frictional energy

The Frictional Energy can produce heat and light energy.

The sandpaper produces little shreds from the grit of the paper.

The shreds have KE (they move)

6 0

2 years ago

Suppose you first walk 12.5 m in a direction 20° west of north and then 24 m in a direction 40° south of west as shown in the fi

Firlakuza [10]

La ce materie vr ca sunt pe tableta

3 0

3 years ago

In an inkjet printer, letters and images are created by squirting drops of ink horizontally at a sheet of paper from a rapidly m

Serga [27]

Answer:

$q = 6.48 \times 10^{-14} C$

Explanation:

Deflection in the drop is due to electric field force

so we will have

$F = qE$

acceleration of the drop is given as

$a = \frac{qE}{m}$

$a = \frac{q(7.75 \times 10^4)}{1.00 \times 10^{-11}}$

$a = 7.75 \times 10^{15} q$

now we know that time to cross the plates is given as

$t = \frac{D}{v}$

$t = \frac{0.02}{18}$

$t = 1.11 \times 10^{-3} s$

now the deflection is given as

$d = \frac{1}{2}at^2$

$0.310 \times 10^{-3} = \frac{1}{2}(7.75 \times 10^{15} q)(1.11 \times 10^{-3})^2$

$0.310 \times 10^{-3} = 4.78 \times 10^9 q$

$q = 6.48 \times 10^{-14} C$

5 0

3 years ago

A particle executes simple harmonic motion with an amplitude of 1.69 cm. At what positive displacement from the midpoint of its

m_a_m_a [10]

Answer: 0.0146m

Explanation: The formula that defines the velocity of a simple harmonic motion is given as

v = ω√A² - x²

Where v = linear velocity, A = amplitude = 1.69cm = 0.0169m, x = displacement.

The maximum speed of a simple harmonic motion is derived when x = A, hence v = ωA

One half of maximum speed = speed of motion

3ωA/2 = ω√A² - x²

ω cancels out on both sides of the equation, hence we have that

A/2 = √A² - x²

(0.0169)/2 = √(0.0169² - x²)

0.00845 = √(0.0169² - x²)

By squaring both sides, we have that

0.00845² = 0.0169² - x²

x² = 0.0169² - 0.00845²

x² = 0.0002142

x = √0.0002142

x = 0.0146m

5 0

3 years ago