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EleoNora [17]
2 years ago
15

A shell is fired from the ground with an initial speed of 1.60 × 103 m/s (approximately five times the speed of sound) at an ini

tial angle of 51.0 ◦ to the horizontal. The acceleration of gravity is 9.81 m/s 2 . a) Neglecting air resistance, find the shell’s horizontal range. Answer in units of m.
Physics
1 answer:
Volgvan2 years ago
6 0

Answer:

The horizontal range will be 2.55\times 10^5m

Explanation:

We have given initial speed of the shell u = 1.6\times 10^3m/sec

Angle of projection = 51°

Acceleration due to gravity g=9.8m/sec^2

We have to find maximum range

Horizontal range in projectile motion is given by

R=\frac{u^2sin2\Theta }{g}=\frac{(1.60\times 10^3)^2sin(2\times 51^{\circ})}{9.81}=2.55\times 10^5m

So the horizontal range will be 2.55\times 10^5m

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Do atoms ever touch​
Mumz [18]

Answer:

nope never

Explanation:

8 0
3 years ago
Read 2 more answers
1) A uniform wooden beam, with mass of 120 and length L = 4 m, is supported as illustrated in the figure. If the static friction
Kobotan [32]

Answer:

1(a) 55.0°

1(b) 58.3°

2(a) 10.2 N

2(b) 2.61 m/s²

3(a) 76.7°

3(b) 12.8 m/s

3(c) 3.41 s

3(d) 21.8 m/s

3(e) 18.5 m

4(a) 7.35 m/s²

4(b) 31.3 m/s²

4(c) 12.8 m/s²

Explanation:

1) Draw a free body diagram on the beam.  There are five forces:

Weight force mg pulling down at the center of the beam,

Normal force Na pushing up at point A,

Friction force Na μa pushing left at point A,

Normal force Nb pushing perpendicular to the incline at point B,

Friction force Nb μb pushing up the incline at point B.

There are 3 unknown variables: Na, Nb, and θ.  So we're going to need 3 equations.

Sum of forces in the x direction:

∑F = ma

-Na μa + Nb sin φ − Nb μb cos φ = 0

Nb (sin φ − μb cos φ) = Na μa

Nb / Na = μa / (sin φ − μb cos φ)

Sum of forces in the y direction:

∑F = ma

Na + Nb cos φ + Nb μb sin φ − mg = 0

Na = mg − Nb (cos φ + μb sin φ)

Sum of torques about point B:

∑τ = Iα

-mg (L/2) cos θ + Na L cos θ − Na μa L sin θ = 0

mg (L/2) cos θ = Na L cos θ − Na μa L sin θ

mg cos θ = 2 Na cos θ − 2 Na μa sin θ

mg = 2 Na − 2 Na μa tan θ

Substitute:

Na = 2 Na − 2 Na μa tan θ − Nb (cos φ + μb sin φ)

0 = Na − 2 Na μa tan θ − Nb (cos φ + μb sin φ)

Na (1 − 2 μa tan θ) = Nb (cos φ + μb sin φ)

1 − 2 μa tan θ = (Nb / Na) (cos φ + μb sin φ)

2 μa tan θ = 1 − (Nb / Na) (cos φ + μb sin φ)

Substitute again:

2 μa tan θ = 1 − [μa / (sin φ − μb cos φ)] (cos φ + μb sin φ)

tan θ = 1/(2 μa) − (cos φ + μb sin φ) / (2 sin φ − 2 μb cos φ)

a) If φ = 70°, then θ = 55.0°.

b) If φ = 90°, then θ = 58.3°.

2) Draw a free body diagram of each mass.  For each mass, there are four forces.  For mass A:

Weight force Ma g pulling down,

Normal force Na pushing perpendicular to the incline,

Friction force Na μa pushing parallel down the incline,

Tension force T pulling parallel up the incline.

For mass B:

Weight force Mb g pulling down,

Normal force Nb pushing perpendicular to the incline,

Friction force Nb μb pushing parallel up the incline,

Tension force T pulling up the incline.

There are four unknown variables: Na, Nb, T, and a.  So we'll need four equations.

Sum of forces on A in the perpendicular direction:

∑F = ma

Na − Ma g cos θ = 0

Na = Ma g cos θ

Sum of forces on A up the incline:

∑F = ma

T − Na μa − Ma g sin θ = Ma a

T − Ma g cos θ μa − Ma g sin θ = Ma a

Sum of forces on B in the perpendicular direction:

∑F = ma

Nb − Mb g cos φ = 0

Nb = Mb g cos φ

Sum of forces on B down the incline:

∑F = ma

-T − Nb μb + Mb g sin φ = Mb a

-T − Mb g cos φ μb + Mb g sin φ = Mb a

Add together to eliminate T:

-Ma g cos θ μa − Ma g sin θ − Mb g cos φ μb + Mb g sin φ = Ma a + Mb a

g (-Ma (cos θ μa + sin θ) − Mb (cos φ μb − sin φ)) = (Ma + Mb) a

a = -g (Ma (cos θ μa + sin θ) + Mb (cos φ μb − sin φ)) / (Ma + Mb)

a = 2.61 m/s²

Plug into either equation to find T.

T = 10.2 N

3i) Given:

Δx = 3.7 m

vᵧ = 0 m/s

aₓ = 0 m/s²

aᵧ = -10 m/s²

t = 1.25 s

Find: v₀ₓ, v₀ᵧ

Δx = v₀ₓ t + ½ aₓ t²

3.7 m = v₀ₓ (1.25 s) + ½ (0 m/s²) (1.25 s)²

v₀ₓ = 2.96 m/s

vᵧ = aᵧt + v₀ᵧ

0 m/s = (-10 m/s²) (1.25 s) + v₀ᵧ

v₀ᵧ = 12.5 m/s

a) tan θ = v₀ᵧ / v₀ₓ

θ = 76.7°

b) v₀² = v₀ₓ² + v₀ᵧ²

v₀ = 12.8 m/s

3ii) Given:

Δx = D cos 57°

Δy = -D sin 57°

v₀ₓ = 2.96 m/s

v₀ᵧ = 12.5 m/s

aₓ = 0 m/s²

aᵧ = -10 m/s²

c) Find t

Δx = v₀ₓ t + ½ aₓ t²

D cos 57° = (2.96 m/s) t + ½ (0 m/s²) t²

D cos 57° = 2.96t

Δy = v₀ᵧ t + ½ aᵧ t²

-D sin 57° = (12.5 m/s) t + ½ (-10 m/s²) t²

-D sin 57° = 12.5t − 5t²

Divide:

-tan 57° = (12.5t − 5t²) / 2.96t

-4.558t = 12.5t − 5t²

0 = 17.058t  − 5t²

t = 3.41 s

d) Find v

vₓ = aₓt + v₀ₓ

vₓ = (0 m/s²) (3.41 s) + 2.96 m/s

vₓ = 2.96 m/s

vᵧ = aᵧt + v₀ᵧ

vᵧ = (-10 m/s²) (3.41 s) + 12.5 m/s

vᵧ = -21.6 m/s

v² = vₓ² + vᵧ²

v = 21.8 m/s

e) Find D.

D cos 57° = 2.96t

D = 18.5 m

4) Given:

R = 90 m

d = 140 m

v₀ = 0 m/s

at = 0.7t m/s²

The distance to the opposite side of the curve is:

140 m + (90 m) (π/2) = 281 m

a) Find Δx and v if t = 10.5 s.

at = 0.7t

Integrate:

vt = 0.35t² + v₀

vt = 0.35 (10.5)²

vt = 38.6 m/s

Integrate again:

Δx = 0.1167 t³ + v₀ t + x₀

Δx = 0.1167 (10.5)³

Δx = 135 m

The car has not yet reached the curve, so the acceleration is purely tangential.

at = 0.7 (10.5)

at = 7.35 m/s²

b) Find Δx and v if t = 12.2 s.

at = 0.7t

Integrate:

vt = 0.35t² + v₀

vt = 0.35 (12.2)²

vt = 52.1 m/s

Integrate again:

Δx = 0.1167 t³ + v₀ t + x₀

Δx = 0.1167 (12.2)³

Δx = 212 m

The car is in the curve, so it has both tangential and centripetal accelerations.

at = 0.7 (12.2)

at = 8.54 m/s²

ac = v² / r

ac = (52.1 m/s)² / (90 m)

ac = 30.2 m/s²

a² = at² + ac²

a = 31.3 m/s²

c) Given:

Δx = 187 m

v₀ = 0 m/s

at = 3 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (0 m/s)² + 2 (3 m/s²) (187 m)

v = 33.5 m/s

ac = v² / r

ac = (33.5 m/s)² / 90 m

ac = 12.5 m/s²

a² = at² + ac²

a = 12.8 m/s²

5 0
3 years ago
A snail crawls 300 cm in 1 hour. Calculate the snail's speed in each of the following units. A) cm/h B) cm/min C) m/h
Crazy boy [7]
A) 300cm/h
B)1 hr=60 min
300/60=5
5cm/min
C)1m=100cm
300/100=3
3m/h
7 0
3 years ago
You have been assigned to investigate a traffic accident. The masses of car A and car B are 1300 kg and 1200 kg, respectively. C
jarptica [38.1K]

Answer:

The velocity of A before impact = 17.90 m/s

Explanation:

Coefficient of restitution = (speed of seperation)/(speed of approach)

= (v₁ - v₂)/(u₂ - u₁)

where v₁ = velocity of the car A after the impact = ?

v₂ = velocity of the car B after the impact = ?

u₂ = velocity of the car B before the impact = 0 m/s (it was initially at rest)

u₁ = velocity of car A before the impact = ?

First of, we can solve for v₂, the velocity of car B after the impact, from some of the information given in the question.

- Skid marks indicate car B slid 10 m after the impact

- The coefficient of kinetic friction the tires and road is 0.8.

According to the work energy theorem, the work done by frictional force in stopping the car B is equal to the change in kinetic energy of the car B. (All after collision)

W = ΔK.E

ΔK.E = (1/2)(1200)(v₂²) - 0 (final kinetic energy is 0 since the car comes to stop eventually)

ΔK.E = (600v₂²) J

W = F × d

where F = frictional force = μmg = 0.8×1300×9.8 = 10,192 N

d = distance the car skids over before stopping = 10 m

W = 10,192 × 10 = 101,920 J

W = ΔK.E

101,920 = 600v₂²

v₂² = (101920/600) = 169.867

v₂ = 13.03 m/s

But recall,

Coefficient of restitution = (v₁ - v₂)/(u₂ - u₁)

For the sake of convention, we take the direction of car A's initial velocity to be the positive direction.

u₁ = ?

u₂ = 0 m/s

v₁ = ?

v₂ = +13.03 m/s

Coefficient of restitution = 0.4

0.4 = (v₁ - 13.03)/(0 - u₁)

-0.4u₁ = v₁ - 13.03

v₁ = 13.03 - 0.4u₁

But this is a collision. In a collision, the linear momentum is usually conserved.

Momentum before collision = Momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1300u₁ + (1200×0) = 1300v₁ + (1200×13.03)

1300u₁ + 0 = 1300v₁ + 15639.95

1300u₁ = 1300v₁ + 15639.95

But recall, from the coefficient of restitution relation,

v₁ = 13.03 - 0.4u₁

Substituting this into the momentum balance equation.

1300u₁ = 1300v₁ + 15639.95

1300u₁ = 1300(13.03 - 0.4u₁) + 15639.95

1300u₁ = 16943.28 - 520u₁ + 15639.95

1820u₁ = 32,583.23

u₁ = (32,583.23/1820)

u₁ = 17.90 m/s

Therefore, the velocity of A before impact = 17.90 m/s

Hope this Helps!!!

4 0
3 years ago
On Mars gravity is one-third that on Earth. What would be the mass on Mars of a person who has a mass of 90 kilograms (kg) on Ea
snow_tiger [21]

Answer: The person will still have a mass of 90kg on Mars

Explanation: The Truth is, the mass of a body remains constant from place to place. It is the weight which is equal to {mass of body * acceleration due to gravity{g}} that varies from place to place since it is dependent on {g}.

In this case the person will have a Weight of 90*9.8 = 882N on Earth.

{ "g" on Earth is 9.8m/s²}

And a Weight of 90*3.3 = 297N on Mars.

{ From the question "g" on Mars is {9.8m/s²}/3 which is 3.3m/s²}

From this analysis you notice that the WEIGHT of the person Varies but the MASS remained Constant at 90kg.

4 0
2 years ago
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