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N76 [4]
3 years ago
11

A plane travels 27,000 meters in 90 seconds. What was its speed?

Physics
2 answers:
Anna11 [10]3 years ago
7 0

Answer:

The answer should be 300 mps.

Explanation:

If you divide 27,000 by 90, you will get the speed traveled by second.

trapecia [35]3 years ago
3 0
The plane goes 300mph
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When a slice of buttered toast is accidentally pushed over the edge of a counter, it rotates as it falls. If the distance to the
Ksenya-84 [330]

Answer: 4.83rad/s and 14.50 rad/s

Explanation:

The distance from the center to the floor d = 52cm = 0.52m

Rotation is less than 1 rev

The toast rotates at a constant angular speed

Using kinematic d = Vit + 1/2gt^2

d = 0+1/2gt^2

t = (2d/g) square root

t = square root of 2× 0.52/ 9.8

t = 0.325s

The toast is accidentally pushed over the edge of the centre with butter side up, then the toast rotates as it falls . If the toast hits the ground and topples, the smallest angle will be less than 1/4 rev with correspondence to the smallest angular speed

Wmin = change in tetha/ change in time

= 0.25 rev/ change in t

= 0.25×2pii / change in t

= 0.5pii/0.325

= 0.5×3.142/0.325

= 4.83 rad/s

The toast is accidentally pushed over the edge of the centre with butter side up, then the toast rotates as it falls . If the toast hits the ground and topples, the maximum angle will be less than 3/4 rev with correspondence to the maximum angular speed

Wmax = change in tetha/ change in time

= 0.75 rev/ Change in time

= 0.75 ×2pii/change in tetha

= 1.5pii/0.325

= 1.5 ×3.142/0.325

= 14.50 rad/s

Note the following

Vi is the initial speed of the toast which is zero because it was initially at rest

t is the time taken for the toast to hit the floor

g is the gravitational acceleration

d is the distance between the counter and the floor. There is an attachment for you as well.

Thanks

5 0
2 years ago
Describe the possible long term impact of a catastrophic event
denpristay [2]
However, the Northridge earthquake itself, a significant catastrophic event<span>, had ... prices, reflecting their increased </span>potential<span> for physical destruction. ... local neighborhoods and find no </span>long-term impact<span> on home values, resident incomes or occupancy rates. ... We briefly </span>describe<span> the variety of data sources used in the paper.</span>
4 0
3 years ago
In a 50 km/h head-on crash test, the steering column of passenger car 1 moved 3 cm upward and 2 cm rearward. The steering column
Softa [21]

Answer:

Car 1

Explanation:

The steering column which moves the least is less likely to to the driver's chest ordinarily. Driver tends to remain in motion until restrained. Assuming a  seat belt not airbag

Generally one would compute a vector find direction and distance.  This is like solving for a hypotenuse  / in a right angled triangle problem. On face value the column moving the least is safer. The 6/24 would hit the upper chest, face, or possibly break the neck.

hence, car 1 moved 3 cm upward and 2 cm rearward is safer.

4 0
3 years ago
An uncharged conductor has a hollow cavity inside of it. Within this cavity there is a charge of +10 µC that does not touch the
aliina [53]

Answer:

Explanation:

we have to make charge inside the conductor zero because we know that electric field inside the conductor should be zero

so,  the outer surface of the conductor should contain + 10 uC of charge and the inner surface contains -10 uC

8 0
3 years ago
A 4.25 kg block is sent up a ramp inclined at an angle theta=37.5° from the horizontal. It is given an initial velocity ????0=15
wel

Answer:

d = 11.79 m

Explanation:

Known data

m=4.25 kg  : mass of the block

θ =37.5°  :angle θ of the ramp with respect to the horizontal direction

μk= 0.460  : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the block

W: Weight of the block : In vertical direction

N : Normal force : perpendicular to the ramp

f : Friction force: parallel to the ramp

Calculated of the W

W= m*g

W=  4.25 kg* 9.8 m/s² = 41.65 N

x-y weight components

Wx= Wsin θ= 41.65*sin 37.5° = 25.35 N

Wy= Wcos θ =41.65*cos 37.5° =33.04 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay    ay = 0

N - Wy = 0

N = Wy

N = 33.04 N

Calculated of the f

f = μk* N= 0.460*33.04

f = 15.2 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

-Wx-f = m*a

 -25.35-15.2 = (4.25)*a

-40.55 =  (4.25)*a

a = (-40.55)/ (4.25)

a = -9.54 m/s²

Kinematics of the block

Because the block moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d Formula (2)

Where:  

d:displacement  (m)

v₀: initial speed  (m/s)

vf: final speed   (m/s)

Data:

v₀ = 15 m/s

vf = 0

a = -9.54 m/s²

We replace data in the formula (2)  to calculate the distance along the ramp the block reaches before stopping (d)

vf²=v₀²+2*a*d

0 = (15)²+2*(-9.54)*d

2*(9.54)*d =   (15)²

(19.08)*d = 225

d = 225 / (19.08)

d = 11.79 m

3 0
3 years ago
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