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Hunter-Best [27]

2 years ago

15

The velocity component with which a projectile covers certain horizontal distance is maximum at the moment of? a) Hitting the gr

ound b) Highest point c) Projection d) None of these

1 answer:

Crazy boy [7]2 years ago

3 0

A) hitting the ground

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A 36,287 kg truck has a momentum of 907,175 kg • m/s. What is the trucks velocity.

Airida [17]

Momentum = Mass x Velocity

Put the values where they belong and solve for Velocity.

In this case, since Mass is being multiplied by Velocity, to solve for be Velocity you would divide both sides by Velocity. The velocity will equal the momentum divided by the mass.

4 0

2 years ago

Slick Willy is in traffic court (again) contesting a $50.00 ticket for running a red light. "You see, your Honor, as I was appro

Masteriza [31]

Answer:

61578948 m/s

Explanation:

λ $_{actual}$ = λ $_{observed}$ $\frac{c+v_{o}}{c}$

687 = 570 $(\frac{3 * 10^{8} +v_{o} }{3 * 10^{8}} )$

$v_{o}$ = 61578948 m/s

So Slick Willy was travelling at a speed of 61578948 m/s to observe this.

8 0

2 years ago

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A seasoned mini golfer is trying to make par on a tricky number five hole. The golfer can complete the hole by hitting the ball

liberstina [14]

Answer:5.17 m/s

Explanation:

Given

let u be the speed at cliff initial point

range over cliff is 1.45 m

and range of projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

$1.45=\frac{u^2\sin 90}{9.8}$

u=3.77 m/s

Conserving Energy

$E_{bottom}=E_{initial\ point\ at\ cliff}$

Kinetic energy=Kinetic energy +Potential energy gained

Let v be the initial velocity

$\frac{mv^2}{2}=mgh+\frac{mu^2}{2}$

$v^2=u^2+2gh$

$v=\sqrt{u^2+2gh}$

$v=\sqrt{3.77^2+2\time 9.8\times 0.64}$

$v=\sqrt{26.75}=5.17 m/s$

5 0

2 years ago

A square loop of wire with initial side length 10 cm is placed in a magnetic field of strength 1 T. The field is parallel to the

Fofino [41]

Answer:

2 x 10⁻³ volts

Explanation:

B = magnetic of magnetic field parallel to the axis of loop = 1 T

$\frac{dA}{dt}$ = rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²

θ = Angle of the magnetic field with the area vector = 0

E = emf induced in the loop

Induced emf is given as

E = B $\frac{dA}{dt}$

E = (1) (20 x 10⁻⁴ )

E = 2 x 10⁻³ volts

E = 2 mV

7 0

2 years ago

A 3.0-kg brick rests on a perfectly smooth ramp inclined at 34° above the horizontal. The brick is kept from sliding down the pl

Firdavs [7]

Answer:

d=0.137 m ⇒13.7 cm

Explanation:

Given data

m (Mass)=3.0 kg

α(incline) =34°

Spring Constant (force constant)=120 N/m

d (distance)=?

Solution

F=mg

F=(3.0)(9.8)

F=29.4 N

As we also know that

Force parallel to the incline=FSinα

F=29.4×Sin(34)

F=16.44 N

d(distance)=F/Spring Constant

d(distance)=16.44/120

d(distance)=0.137 m ⇒13.7 cm

4 0

2 years ago