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Reika [66]
3 years ago
14

A truck traveling at 100 km/hr strikes an unfortunate bug and splatters it. The acceleration resulting from the impact is:______

_
a. Greater for the truck
b. Greater for the bug
b. The same for both the bug and the truck
d. Impossible to determine
Physics
1 answer:
KATRIN_1 [288]3 years ago
4 0

Answer:

<h2>b. Greater for the bug</h2>

Explanation:

The conservation of linear momentum states that the momentum before impact equals the momentum after impact

Also, when two objects collide,  the change of momentum of particles m1 and m2 are equal in magnitude and opposite in sign and the total momentum change equals zero. Hence during a collision, Each object experiences the same force for the same amount of time, leading to the same impulse, and subsequently the same momentum change. in general, it is only the acceleration and the velocity components that differ for the two objects during the interaction.

Consequently, The object with the least mass always receives the greatest velocity change and acceleration.

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A rectangular coil with 50 turns of conducting wire and a total resistance of 10.0 Ω initially lies in the yz-plane at time t =
castortr0y [4]

Answer:

a) 43.20V

b) 2.71W/s

c) 40.25s

d) 7.77Nm

Explanation:

(a) The emf of a rotating coil with N turns is given by:

emf=NBA\omega sin(\omega t)

N: turns

B: magnitude of the magnetic field

A: area

w: angular velocity

the emf max is given by:

emf_{max}=NBA\omega=(50)(1.80T)(0.200m*0.100m)(24.0rad/s)\\\\emf_{max}=43.20V

(b) the maximum rate of change of the magnetic flux is given by:

\frac{d\Phi_B}{dt}=\frac{d(A\cdot B)}{dt}=\frac{d}{dt}(ABcos\omega t)=AB\omega sin(\omega t)\\\\\frac{d\Phi_B}{dt}_{max}=(\pi(0.200*0.100))(1.80T)(24.0rad/s)=2.71\frac{W}{s}

(c) emf(t=0.050s)=(50)(1.80T)(0.200m*0.100m)(24rad/s)sin(24.0rad/s(0.050s))\\\\emf(t=0.050s)=40.26V

(d) The torque is given by:

\tau=NABIsin\theta\\\\NAB\omega=emf_{max}\\\\\tau=\frac{emf_{max}}{\omega}\frac{emf_{max}}{R}\\\\\tau=\frac{(43.20V)^2}{(24.0rad/s)(10.0\Omega)}=7.77Nm

3 0
3 years ago
Which of these accurately identify characteristics of light and mechanical waves? Select the TWO (2) that apply.
Tanzania [10]

The two correct statement are A and B. Light waves are electromagnetic waves that can travel through a vacuum. Mechanical waves can travel through a vacuum.

<h3>What is an electromagnetic wave?</h3>

EM waves are the electromagnetic radiations are made up of electromagnetic waves created when an electric field collides with a magnetic field.

Electromagnetic waves may alternatively be defined as the combination of oscillating electric and magnetic fields.

Electrically charged particles experiencing acceleration create electromagnetic waves, which can then interact with other charged particles and exert force on them.

The two correct statement is;

1. Light waves are electromagnetic waves that can travel through a vacuum.

2. Mechanical waves can travel through a vacuum.

Hence,two correct statement are A and B.

To learn more about the electromagnetic wave refer to the link;

brainly.com/question/8553652

#SPJ1

4 0
2 years ago
Mia wanted to know how Earth’s movements created new landforms. She decided to read about folded mountains and their characteris
kifflom [539]

Answer:

I’m pretty sure it’s D

Explanation:

the crust breaks and shifts when that happens

hope dis helps ^-^

5 0
3 years ago
Read 2 more answers
A plastic rod that has been charged to -21 nc touches a metal sphere. afterward, the rod's charge is -8 nc. (a) what kind of cha
otez555 [7]
The rod transferred negatively charged particle to the metal sphere. The rod becomes less negatively charged than it was originally (originally it was charged to -21 nc, now it has -8, and -8>-21), so it must have moved 13 nc of negative charge to the sphere. 
3 0
4 years ago
The maximum lift-to-drag ratio of the World War I Sopwith Camel was 7.7. If the aircraft is in flight at 5000 ft when the engine
andre [41]

The related concepts to solve this problem is the Glide Ratio. This can be defined as the product between the height of fall and the lift-to-drag ratio. Mathematically, this expression can be written as,

R = h (\frac{L}{D})_{max}

Replacing,

R = 5000ft (7.7)

R = 38500ft

Converting this units to miles.

R = 38500ft (\frac{1mile}{5280ft})

R = 7.2916miles

Therefore the glide in terms of distance measured along the ground is 7.2916miles

3 0
4 years ago
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