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3 years ago

If the fly experienced a force of 100 N when it hits how big was the force on the vehicle in this collision

Physics

1 answer:

iogann1982 [59]3 years ago

8 0

Answer:

100N

Explanation:

because newton's third law of motion states that if body A exerts a force on the body B, then Body B will exert an equal force but opposite in direction force on body A

so if the fly experienced 100N, then the car will also experience 100N

hope you get it

please mark

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An object at rest starts accelerating.

Shalnov [3]

Answer:

We are given:

initial velocity (u) = 0 m/s

final velocity (v) = 10 m/s

displacement (s) = 20 m

acceleration (a) = a m/s/s

Solving for 'a'

From the third equation of motion:

v² - u² = 2as

replacing the variables

(10)² - (0)² = 2(a)(20)

100 = 40a

a = 100 / 40

a = 2.5 m/s²

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3 years ago

If a 375 watt heater has a current of 5.0 A, what is the resistance of the heating element?

liberstina [14]

P=IV
V=IR

P=I(IR)
P=I²R
375=5²R
R=375/25
R=15

7 0

3 years ago

The interior space of large box is kept at 30 C. The walls of the box are 3 m high and have a ‘sandwich’ construction consisting

White raven [17]

Answer:

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$

$T_1=27.58\ ^{\circ}C$ & $T_2=2.41875\ ^{\circ}C$

Explanation:

Given:

interior temperature of box, $T_i=30^{\circ}C$

height of the walls of box, $h=3\ m$

thickness of each layer of bi-layered plywood, $x_p=1.25\ cm=0.0125\ m$

thermal conductivity of plywood, $k_p=0.104\ W.m^{-1}.K^{-1}$

thickness of sandwiched Styrofoam, $x_s=5\ cm=0.05\ m$

thermal conductivity of Styrofoam, $k_s=0.04\ W.m^{-1}.K^{-1}$

exterior temperature, $T_o=0^{\circ}C$

From the Fourier's law of conduction:

$\dot Q=\frac{dT}{(\frac{x}{kA}) }$

$\dot Q=\frac{dT}{R_{th} }$ ....................................(1)

Now calculating the equivalent thermal resistance for conductivity using electrical analogy:

$R_{th}=R_p+R_s+R_p$

$R_{th}=\frac{x_p}{k_p.A}+\frac{x_s}{k_s.A}+\frac{x_p}{k_p.A}$

$R_{th}=\frac{1}{A} (\frac{x_p}{k_p}+\frac{x_s}{k_s}+\frac{x_p}{k_p})$

$R_{th}=\frac{1}{A} (\frac{0.0125}{0.104}+\frac{0.05}{0.04}+\frac{0.0125}{0.104})$

$R_{th}=\frac{1.4904}{A}$ .....................(2)

Putting the value from (2) into (1):

$\dot Q=\frac{30-0}{\frac{1.4904}{A} }$

$\dot Q=\frac{30\ A}{1.4904}$

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$ is the heat per unit area of the wall.

The heat flux remains constant because the area is constant.

For plywood-Styrofoam interface from inside:

$\frac{\dot Q}{A} =k_p.\frac{T_i-T_1}{x_p}$

$20.129=0.104\times \frac{30-T_1}{0.0125}$

$T_1=27.58\ ^{\circ}C$

&For Styrofoam-plywood interface from inside:

$\frac{\dot Q}{A} =k_s.\frac{T_1-T_2}{x_s}$

$20.129=0.04\times \frac{27.58-T_2}{0.05}$

$T_2=2.41875\ ^{\circ}C$

4 0

3 years ago

A 3500 N is force is applied to a spring that has a spring of constant of k= 14000 N/m. How far from equilibrium will the spring

Taya2010 [7]

Answer:

the spring be displaced by 25.0 cm

Explanation:

The computation is shown below:

As we know that

F= -K × x

So,

$x = \frac{-F}{K}$

Now

$x = \frac{-3500}{14000} \\\\$

= -0.250m

= 25.0 cm

Hence, the spring be displaced by 25.0 cm

4 0

3 years ago

A quantity found by multiplying the force by the distance moved

Andrej [43]

The quantity that is calculated from the product of the force and the distance traveled due to the force is called work. It has SI units of Joules (J) which is equivalent to Newton-meter (N-m). It is the energy that happens when an object is being moved by an external force.

7 0

3 years ago