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3 years ago

PLS ANSWER FAST WILL GIVE BRAINLEST!!!

Physics

1 answer:

Kipish [7]3 years ago

8 0

Answer:

2 Newtons

Explanation:

$F = ma$

Therefore, your mass would be 1kg and your acceleration would be 2m/s/s

Plug the numbers into the equation:

$(1kg)(2m/s/s)$

which will equal

$2 Newtons$

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Now let’s apply the work–energy theorem to a more complex, multistep problem. In a pile driver, a steel hammerhead with mass 200

andrew11 [14]

Answer:

a) v = 7.67

b) n = 81562 N

Explanation:

Given:-

- The mass of hammer-head, m = 200 kg

- The height at from which hammer head drops, s12 = 3.00 m

- The amount of distance the I-beam is hammered, s23 = 7.40 cm

- The resistive force by contact of hammer-head and I-beam, F = 60.0 N

Find:-

(a) the speed of the hammerhead just as it hits the I-beam and

(b) the average force the hammerhead exerts on the I-beam.

Solution:-

- We will consider the hammer head as our system and apply the conservation of energy principle because during the journey of hammer-head up till just before it hits the I-beam there are no external forces acting on the system:

ΔK.E = ΔP.E

K_2 - K_1 = P_1- P_2

Where, K_2: Kinetic energy of hammer head as it hits the I-beam

K_1: Initial kinetic energy of hammer head ( = 0 ) ... rest

P_2: Gravitational potential energy of hammer head as it hits the I-beam. (Datum = 0)

P_1: Initial gravitational potential energy of hammer head

- The expression simplifies to:

K_2 = P_1

Where, 0.5*m*v2^2 = m*g*s12

v2 = √(2*g*s12) = √(2*9.81*3)

v2 = 7.67 m/s

- For the complete journey we see that there are fictitious force due to contact between hammer-head and I-beam the system is no longer conserved. All the kinetic energy is used to drive the I-beam down by distance s23. We will apply work energy principle on the system:

Wnet = ( P_3 - P_1 ) + W_friction

Wnet = m*g*s13 + F*s23

n*s23 = m*g*s13 + F*s23

Where, n: average force the hammerhead exerts on the I-beam.

s13 = s12 + s23

Hence,

n = m*g*( s12/s23 + 1) + F

n = 200*9.81*(3/0.074 + 1) + 60

n = 81562 N

6 0

3 years ago

An escalator is used to move 25 passengers every minute from the first floor of a department store to the second. The second flo

kompoz [17]

Answer:

55960 J

Explanation:

8 0

3 years ago

Read 2 more answers

An important dimensionless parameter in certain types of fluid flow problems is the Froude number defined as V/√g.l where V is a

prohojiy [21]

Answer:

1.24611

Explanation:

V = Velocity = 10 ft/s

L = Length = 2 ft

g = Acceleration due to gravity = 32.2 ft/s²

Froude number is given by

$Fr=\dfrac{V}{\sqrt{gL}}\\\Rightarrow Fr=\dfrac{10}{\sqrt{32.2\times 2}}\\\Rightarrow Fr=1.24611$

Converting to SI units

$10\ ft/s=10\times \dfrac{1}{3.281}$

$32.2\ ft/s^2=32.2\times \dfrac{1}{3.281}$

$2\ ft=2\times \dfrac{1}{3.281}$

$Fr=\dfrac{V}{\sqrt{gL}}\\\Rightarrow Fr=\dfrac{10\times \dfrac{1}{3.281}}{\sqrt{32.2\times \dfrac{1}{3.281}\times 2\times \dfrac{1}{3.281}}}\\\Rightarrow Fr=1.24611$

The Froude number is 1.24611

The Froude number is equal. The Froude number is dimensionless as the units cancel each other. In order for this to happen the units used need to be consitent either imperial or SI.

7 0

3 years ago

Ideally, the resistance of an ammeter should be:

pav-90 [236]

Ideally the resistance should be ZERO

7 0

3 years ago

Two blocks connected by a rope of negligible mass are being dragged by a horizontal force (see figure below). Suppose F = 65.0 N

vovangra [49]

Refer to the diagram shown below.

g = 9.8 m/s², and air resistance is ignored.

For mass m₁:
The normal reaction is m₁g.
The resisting force is R₁ = μm₁g.

For mass m₂:
The normal reaction is m₂g.
The resisting force is R₂ = μm₂g.

Let a = the acceleration of the system.
Then
(m₁ + m₂)a = F - (R₁ + R₂)
(14+26 kg)*(a m/s²) = (65 N) - 0.098*(9.8 m/s²)*(14+26 kg)
40a = 65 - 38.416 = 26.584
a = 0.6646 m/s²

Answer: 0.665 m/s² (nearest thousandth)

7 0

3 years ago