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iogann1982 [59]

4 years ago

14

Daring Darless wishes to cross the Grand Canyon of the Snake River by being shot from a cannon. She wishes to be launched at 65

∘ relative to the horizontal so she can spend more time in the air waving to the crowd.
With what minimum speed must she be launched to cross the 520-m gap?
Express your answer to two significant figures and include the appropriate units.

1 answer:

guajiro [1.7K]4 years ago

3 0

Answer:

She must be launched with minimum speed of <u>57.67 m/s</u> to clear the 520 m gap.

Step-by-step explanation:

Given:

The angle of projection of the projectile is, $\theta =65$ °

Range of the projectile is, $R=520$ m.

Acceleration due to gravity, $g=9.8\ m/s^2$

The minimum speed to cross the gap is the initial speed of the projectile and can be determined using the formula for range of projectile.

The range of projectile is given as:

$R=\frac{v_{0}^2\sin2\theta}{g}$

Plug in all the given values and solve for minimum speed, $v_0$ .

$520=\frac{v_{0}^2\sin(2(65))}{9.8}\\520\times 9.8=v_{0}^2\sin(130)\\5096=1.532v_{0}^2\\v_0^2=\frac{5096}{1.532}\\v_0^2=3326.371\\v_0=\sqrt{3326.371}=57.67\textrm{ m/s}$

Therefore, she must be launched with minimum speed of 57.67 m/s to clear the 520 m gap.

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In the late 19th century, great interest was directed toward the study of electrical discharges in gases and the nature of so-ca

gizmo_the_mogwai [7]

Answer:

A) He finds the same value of q / m for different materials , B) y = ½ (q / m) E L² / v₀ₓ² , C) v = E / B , D) B = 2.13 10⁻⁶ T, E) For the first part I have two off-center points., For the second part I can center one point but the other is off center

Therefore the third statement is correct

Explanation:

Part A

Thomson's experiments are the first proof that the atoms that until now were considered indivisible were constituted by different elements, in these experiments Thomson himself the ratio q / m of several cathodes and always found the same value, which allowed to establish that In atoms there are two types of particles, some of which are mobile and others are still.

When examining his statements the correct one is: He finds the same value of q / m for different materials

Part B

For this part let's use Newton's second law

F = ma

q E = m a

a = q / m E

We use the kinematic relationship

y = voy t - ½ to t²

x = v₀ₓ t

The initial vertical velocity of electrons is zero

y = ½ a (x / v₀ₓ)²

We replace

y = ½ (q / m) E L² / v₀ₓ²

Part C

If there is no deflection, the electric and magnetic forces are the same and in the opposite direction

Fm = Fe

q v B = q E

v = E / B

Part D

We replace

y = ½ (q / m) E L² / (E / B)²

y = ½ (q / m) L² B² / E

If we do not want any deflection the magnetic field has to return the electrons the amount that they lower y = -4.12 cm

-4.12 10⁻² = ½ q / m 0.12² B² / 1.1 10³

-16.97 10⁻⁴ = 6.54 10⁻³ B² q / m

B² = -2.59 10⁻¹ q / m

q / m = -1.758 10¹¹ C/ kg

B = √ 0.259 1.758 10¹¹ = √ 4.55 10⁻¹²

B = 2.13 10⁻⁶ T

Part E

As the charge that the two particles is different

For the first part I have two off-center points.

For the second part I can center one point but the other is off center

Therefore the third statement is correct

8 0

3 years ago

A horse walks along a straight path. it travels 3 in one second, 4 m in the next second, and 5 m in the third second. describe t

Temka [501]

Answer: The horse is moving with a uniform acceleration

Explanation:

According to the described situation, we are dealing with a <u>constant acceleration</u> (also called <u>uniform acceleration</u>), since the horse's velocity is changingn by a constant rate.

Let's prove it:

Firstly we are told the horse follows a straight path. In addition we are given its velocity:

3 m/s, then 4 m/s and 5 m/s

Since acceleration $a$ is defined as the change of velocity in time we can calculate it:

$a\frac{4 m/s-3 m/s}{1 s}=1m/s^{2}$ This is the horse's constant acceleration

3 0

3 years ago

(a) An ideal gas initially at pressure p0 undergoes a free expansion until its volume is 3.80 times its initial volume. What the

trapecia [35]

Answer:

a)P/Po=0.263

b)γ=1.33 So gas is triatomic

c) $\dfrac{KE_f}{KE_i}=1.56$

Explanation:

a)

initial pressure = Po

Initial volume = Vo

Final volume = 3.8 Vo

Lets take final pressure is P

we know that for free expansion process

PV= Constant

Po x Vo = P x 3.8 Vo

P=0.263 Po

So

P/Po=0.263

b)

Now gas is compressed in adiabatic manner

Final pressure = 1.56 Po

=1.56 Po

We know that for adiabatic process

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

$\dfrac{V_2}{V_1}=\left(\dfrac{P_1}{P_2}\right)^{\dfrac{1}{\gamma}}$

$0.263P_o(3.8V_o)^{\gamma}=1.56P_o\times V_o^{\gamma}$

γ=1.33 So gas is triatomic

c)

We know that average kinetic energy given as

$KE=\dfrac{3}{2}KT$

$\dfrac{KE_f}{KE_i}=\dfrac{T_f}{T_i}$ \

$\dfrac{KE_f}{KE_i}=\dfrac{P_fV_f}{P_iV_i}$

$\dfrac{KE_f}{KE_i}=\dfrac{1.56P_oV_o}{P_oV_o}$

$\dfrac{KE_f}{KE_i}=1.56$

3 0

3 years ago

Four model rockets are launched in a field. The mass of each rocket and the net force acting on it when it launches are given in

kondor19780726 [428]

Answer:

Rocket 2 has highest acceleration.

Explanation:

Net force,

F = mass (m) × acceleration (a)

We have,

m₁ = 4.25 kg and F = 120 N

$a_1=\dfrac{F_1}{m_1}\\\\a_1=\dfrac{120}{4.25}\\\\a_1=28.23\ m/s^2$

m₂ = 3.25 kg, F₂ = 120 N

$a_2=\dfrac{F_2}{m_2}\\\\a_2=\dfrac{120}{3.25}\\\\a_2=36.92\ m/s^2$

m₃ = 5.5 kg, F₃ = 120 N

$a_3=\dfrac{F_3}{m_3}\\\\a_3=\dfrac{120}{5.5}\\\\a_3=21.81\ m/s^2$

m₄ = 4.5 kg, F₄ = 120 N

$a_4=\dfrac{F_4}{m_4}\\\\a_4=\dfrac{120}{4.5}\\\\a_4=26.66\ m/s^2$

Hence, it can be seen that the highest acceleration is of rocket 2.

7 0

3 years ago

A mass of 1 slug is suspended from a spring whose spring constant is 9 lb/ft. The mass is initially released from a point 1 foot

FromTheMoon [43]

Answer:

t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number

Explanation:

To solve the problem/ we first write the differential equation governing the motion. So,

$m\frac{d^{2} x}{dt^{2} } = -kx \\ m\frac{d^{2} x}{dt^{2} } + kx = 0\\\frac{d^{2} x}{dt^{2} } + \frac{k}{m} x = 0$

with m = 1 slug and k = 9 lb/ft, the equation becomes

$\frac{d^{2} x}{dt^{2} } + \frac{9}{1} x = 0\\\frac{d^{2} x}{dt^{2} } + 9 x = 0$

The characteristic equation is

D² + 9 = 0

D = ±√-9 = ±3i

The general solution of the above equation is thus

x(t) = c₁cos3t + c₂sin3t

Now, our initial conditions are

x(0) = -1 ft and x'(0) = -√3 ft/s

differentiating x(t), we have

x'(t) = -3c₁sin3t + 3c₂cos3t

So,

x(0) = c₁cos(3 × 0) + c₂sin(3 × 0)

x(0) = c₁cos(0) + c₂sin(0)

x(0) = c₁ × (1) + c₂ × 0

x(0) = c₁ + 0

x(0) = c₁ = -1

Also,

x'(0) = -3c₁sin(3 × 0) + 3c₂cos(3 × 0)

x'(0) = -3c₁sin(0) + 3c₂cos(0)

x'(0) = -3c₁ × 0 + 3c₂ × 1

x'(0) = 0 + 3c₂

x'(0) = 3c₂ = -√3

c₂ = -√3/3

So,

x(t) = -cos3t - (√3/3)sin3t

Now, we convert x(t) into the form x(t) = Asin(ωt + Φ)

where A = √c₁² + c₂² = √[(-1)² + (-√3/3)²] = √(1 + 1/3) = √4/3 = 2/√3 = 2√3/3 and Ф = tan⁻¹(c₁/c₂) = tan⁻¹(-1/-√3/3) = tan⁻¹(3/√3) = tan⁻¹(√3) = π/3.

Since tanФ > 0, Ф is in the third quadrant. So, Ф = π/3 + π = 4π/3

x(t) = (2√3/3)sin(3t + 4π/3)

So, the velocity v(t) = x'(t) = (2√3)cos(3t + 4π/3)

We now find the times when v(t) = 3 ft/s

So (2√3)cos(3t + 4π/3) = 3

cos(3t + 4π/3) = 3/2√3

cos(3t + 4π/3) = √3/2

(3t + 4π/3) = cos⁻¹(√3/2)

3t + 4π/3 = ±π/6 + 2kπ where k is an integer

3t = ±π/6 + 2kπ - 4π/3

t = ±π/18 + 2kπ/3 - 4π/9

t = π/18 + 2kπ/3 - 4π/9 or -π/18 + 2kπ/3 - 4π/9

t = π/18 - 4π/9 + 2kπ/3 or -π/18 - 4π/9 + 2kπ/3

t = -7π/18 + 2kπ/3 or -π/2 + 2kπ/3

Since t is not less than 0, the values of k ≤ 0 are not included

So when k = 1,

t = 5π/18 and π/6. So,

t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number

6 0

3 years ago