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2 years ago

A girl pushes an 18.15 kg wagon with a force of 3.63 N. what is the acceleration?

Physics

2 answers:

dimulka [17.4K]2 years ago

5 0

Answer:0.2 m/s^2

Explanation:

mass=18.15kg

Force=3.63N

Acceleration=force ➗ mass

Acceleration=3.63 ➗ 18.15

Acceleration=0.2 m/s^2

Anton [14]2 years ago

3 0

Divide the force given by mass and you will find the acceleration of the object :-

F = m × a

3.63 = 18.15 × a

3.63 = 18.15a

a = 3.63/18.15

a = 0.2 m/s^2

hope it helps!

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Two fun-loving otters are sliding toward each other on a muddy (and hence frictionless) horizontal surface. One of them, of mass

zvonat [6]

Answer:

(a). The magnitude and direction of the velocity of the otters after collision is 1.35 m/s toward left.

(b). The mechanical energy dissipates during this play is 226.98 J.

Explanation:

Given that,

Mass of one otter = 8.50 kg

Speed = 6.00 m/s

Mass of other = 5.75 kg

Speed = 5.50 m/s

(a). We need to calculate the magnitude and direction of the velocity of these free-spirited otters right after they collide

Using conservation of momentum

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v$

Put the value into the formula

$8.50\times(-6.00)+5.75\times5.50=(8.50+5.75)\times v$

$v=\dfrac{-19.375}{14.25}$

$v=-1.35\ m/s$

Negative sign shows the direction of motion of the object after collision is toward left.

(b). We need to calculate the initial kinetic energy

Using formula of kinetic energy

$K.E_{i}=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2$

Put the value into the formula

$K.E_{i}=\dfrac{1}{2}\times8.50\times(6.00)^2+\dfrac{1}{2}\times5.75\times(5.50)^2$

$K.E_{i}=239.96\ J$

We need to calculate the final kinetic energy

Using formula of kinetic energy

$K.E_{f}=\dfrac{1}{2}(m_{1}+m_{2})v^2$

Put the value into the formula

$K.E_{f}=\dfrac{1}{2}\times(8.50+5.75)\times(-1.35)^2$

$K.E_{f}=12.98\ J$

We need to calculate the mechanical energy dissipates during this play

Using formula of loss of mechanical energy

$\Delta K.E=K.E_{f}-K.E_{i}$

Put the value into the formula

$\Delta K.E=12.98-239.96$

$\Delta K.E=-226.98\ J$

Negative sign shows the loss of mechanical energy

Hence, (a). The magnitude and direction of the velocity of the otters after collision is 1.35 m/s toward left.

(b). The mechanical energy dissipates during this play is 226.98 J.

8 0

2 years ago

Read 2 more answers

calculate the time rate of change in air density during expiration. Assume that the lung has a total volume of 6000mL, the diame

kipiarov [429]

Answer:

The time rate of change in air density during expiration is 0.01003kg/m³-s

Explanation:

Given that,

Lung total capacity V = 6000mL = 6 × 10⁻³m³

Air density p = 1.225kg/m³

diameter of the trachea is 18mm = 0.018m

Velocity v = 20cm/s = 0.20m/s

dv /dt = -100mL/s (volume rate decrease)

= 10⁻⁴m³/s

Area for trachea =

$\frac{\pi }{4} d^2\\= 0.785\times 0.018^2\\= 2.5434 \times10^-^4m^2$

0 - p × Area for trachea =

$\frac{d}{dt} (pv)=v\frac{ds}{dt} + p\frac{dv}{dt}$

$-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)$

⇒ $-0.623133\times10^-^4+1.225\times10^-^4=6\times10^-^3\frac{ds}{dt}$

$\frac{ds}{dt} = \frac{0.6018\times10^-^4}{6\times10^-^3} \\\\= 0.01003kg/m^3-s$

ds/dt = 0.01003kg/m³-s

Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s

3 0

2 years ago

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A nonlinear spring is used to launch a toy car. The car is pushed against the spring, compressing the spring 2.5 cm. The force t

dybincka [34]

Answer:

1.56 J

Explanation:

given,

Spring compression, x = 2.5 cm

Force exerts by the spring,

F = - k x

k = 5000 N/m

Potential energy stored = ?

energy stored in the spring

$PE = \dfrac{1}{2}kx^2$

$PE = \dfrac{1}{2}\times 5000\times 0.025^2$

PE = 1.56 J

Hence, the potential energy stored in the car is equal to 1.56 J.

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2 years ago

Complete this equation that represents the process of nuclear fission.

ki77a [65]

Answer:

a=146 b=56

Explanation:

edg-2020

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2 years ago

A projectile is launched at an angle of 45° from the horizontal and lands 21 s later at the same height from which it was launch

irinina [24]

Answer:

a) initial speed of projectile = 145.5 m/s

b) Maximum altitude = 540 m

c) Range = 2160.6 m

d) r = (1440î + 480j) m

Explanation:

The distance at any time for the projectile is given by the relation - r² = x² + y²

where x = horizontal distance covered covered by the projectile and y = vertical distance coveredby the projectile

Let the initial velocity be u = ?

angle of projection be θ with respect to the horizontal = 45°

u = (uₓî + uᵧj) m/s

T = total time of flight = 21 s

t = any time during the flight of the projectile

a) Total time of flight = 2 uᵧ/g = (2u sin θ)/g

21 = (2u sin 45°)/9.8

u = 145.5 m/s

b) maximum altitude of the projectile = H

H = (u² sin² θ)/2g

H = (145.5² sin² 45°)/(2 × 9.8)

H = 540 m

c) According to projectile motion the maximum horizontal displacement is given by

x = R = uₓT = u cos(θ) T (since uₓ = u cos θ)

R = (145.5 cos 45°) × 21 = 2160.6 m

d) At 14 s,

x = uₓt = u cos(θ) t (since uₓ = u cos θ)

x = (145.5 cos 45°) × 14 = 1440.1 m

y = uᵧ t - 0.5gt² = [u sin(θ)] t - 0.5gt² = (145.5 sin 45°) × 14 - 0.5(9.8)(14) = 480 m

r = (1440î + 480j) m

6 0

2 years ago