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Anettt [7]
4 years ago
10

A plane travelling at 63 m/s[S] down a runway begins accelerating uniformly at 2.8 m/s?[S]. How far does it travel

Physics
1 answer:
AVprozaik [17]4 years ago
6 0

Answer:

270 m

Explanation:

Given:

v₀ = 63 m/s

a = 2.8 m/s²

t = 4.0 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (63 m/s) (4.0 s) + ½ (2.8 m/s²) (4.0 s)²

Δx = 274.4 m

Rounded to two significant figures, the displacement is 270 meters.

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Willie, in a 100.0 m race, initially accelerates uniformly from rest at 2.00 m/s2 until reaching his top speed of 12.0 m/s. He m
Oduvanchick [21]

Answer:

The total time for the race is 11.6 seconds

Explanation:

The parameters given are;

Total distance ran by Willie = 100.0 m

Initial acceleration = 2.00m/s²

Top speed reached with initial acceleration = 12.0 m/s

Point where Willie start to fade and decelerate = 16.0 m from the finish line

Speed with which Willie crosses the finish line = 8.00 m/s

The time and distance covered with the initial acceleration are found using the following equations of motion;

v = u₀ + a·t

v² = u₀² + 2·a·s

Where:

v = Final velocity reached with the initial acceleration = 12.0 m/s

u₀ = Initial velocity at the start of the race = 0 m/s

t = Time during acceleration

a = Initial acceleration = 2.00 m/s²

s = Distance covered during the period of initial acceleration

From, v = u₀ + a·t, we have;

12 = 0 + 2×t

t = 12/2 = 6 seconds

From, v² = u₀² + 2·a·s, we have;

12² = 0² + 2×2×s

144 = 4×s

s = 144/4 =36 meters

Given that the Willie maintained the top speed of 12.0 m/s until he was 16.0 m from the finish line, we have;

Distance covered at top speed = 100 - 36 - 16 = 48 meters

Time, t_t of running at top speed = Distance/velocity = 48/12 = 4 seconds

The deceleration from top speed to crossing the line is found as follows;

v₁² = u₁² + 2·a₁·s₁

Where:

u₁ = v = 12 m/s

v₁ = The speed with which Willie crosses the line = 8.00 m/s

s₁ = Distance covered during decelerating = 16.0 m

a₁ = Deceleration

From which we have;

8² = 12² + 2 × a × 16

64 = 144 + 32·a

64 - 144 = 32·a

32·a = -80

a = -80/32 = -2.5 m/s²

From, v₁ = u₁ + a₁·t₁

Where:

t₁ = Time of deceleration

We have;

8 = 12 + (-2.5)·t₁

t₁ = (8 - 12)/(-2.5) = 1.6 seconds

The total time = t + t_t + t₁ =6 + 4 + 1.6 = 11.6 seconds.

6 0
4 years ago
A box at rest is in a state of equilibrium half way up on a ramp. The ramp has an incline of 42° . What is the force of static f
olchik [2.2K]

Answer:

Explanation:

The force of static friction acting on the box is the frictional force;

Frictional force Ff = Wsin theta (force acting along the ramp)

W is the gravitational force known as the weight

Ff = 112.1sin42°

Ff = 112.1(0.6691)

Ff = 75.00N

Hence he force of static friction acting on the box  if box has a gravitational force of 112.1 N id 75.00N

4 0
3 years ago
Who has the greater acceleration rate? A runner
Arturiano [62]
The first runner because it is very clear that accelarition depends on the time and we know that the time in this case is pretty simple
6 0
4 years ago
For your senior project, you would like to build a cyclotron that will accelerate protons to 10% of the speed of light. The larg
jeyben [28]

Answer:

<h2>Magnetic field strength in that region is 1.2 T</h2>

Explanation:

As we know by the formula of radius of charge moving in external field is given as

R = \frac{mv}{qB}

so we will have

R = 25 cm

m = 1.6 \times 10^{-27} kg

q = 1.6 \times 10^{-19} C

v = 0.10 \times 3 \times 10^8 m/s

now we have

0.25 = \frac{(1.6 \times 10^{-27})(3\times 10^7)}{(1.6 \times 10^{-19})B}

now we have

B = 1.2 T

8 0
3 years ago
A rifle fires a 2.10 x 10^-2 kg pellet straight upward, because the pellet rests on a compressed spring when the trigger is is p
yKpoI14uk [10]

m = mass of pellet = 2.10 x 10⁻² kg

x = compression of spring at the time launch = 9.10 x 10⁻² m

h = height gained by the pellet above the initial position = 6.10 m

k = spring constant

using conservation of energy

spring potential energy = gravitational potential energy of pellet

(0.5) k x² = m g h

inserting the values

(0.5) k (9.10 x 10⁻²)² = (2.10 x 10⁻²) (9.8) (6.10)

k = 303.2 N/m

4 0
3 years ago
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