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2 years ago

11

An unfortunate 18 kg monkey falls from a 40 m tall tree. What is the monkeys final velocity just befor he impacts the ground.? a

7063.2 m/s b 784 m/s c 28 m/s d 720 m/s

1 answer:

zalisa [80]2 years ago

5 0

The correct answer is c) 28 m/s.
Let's find the step-by-step solution. The motion of the monkey is an uniformly accelerated motion, with acceleration equal to $g=9.81 m/s^2$ . The initial velocity of the monkey is zero, while the distance covered is S=40 m. Therefore, we can use the following relationship to find vf, the final velocity of the monkey:
$2aS=v_f^2-v_i^2=v_f^2$
from which
$v_f= \sqrt{2aS}= \sqrt{2\cdot 9.81 m/s^2 \cdot 40 m}=28 m/s$

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A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann

Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/ $s^{2}$

Bring out the given parameters from the question:

Initial Velocity ( $V_{1}$ ) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) = 800 m

Projection angle ∅ = ?

Horizontal distance = $V_{1x}$ tcos ∅ .......................... Equation 1

where $V_{1x}$ = velocity in the X - direction

t = Time taken

Vertical Distance = y = $V_{1y}$ t - $\frac{1}{2}$ g $t^{2}$ ................... Equation 2

Where $V_{1y}$ = Velocity in the Y- direction

t = Time taken

$V_{1y}$ = $V_{1}$ sin∅

Making time (t) subject of the formula in Equation 1

t = d/( $V_{1x}$ cos ∅)

t = $\frac{2000}{1000coso}$ = $\frac{2}{cos0}$ = $\frac{d}{cos o}$ ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = $V_{1y}$ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = sin∅ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = dtan∅ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Applying geometry

$\frac{1}{cos o}$ = $tan^{2} o$ + 1

Vertical Distance = h = d tan∅ - 2 g ( $tan^{2} o$ + 1)

substituting the given parameters

800 = 2000 tan ∅ - 2 (9.81)( $tan^{2} o$ + 1)

800 = 2000 tan ∅ - 19.6( $tan^{2} o$ + 1) Equation 4

Replacing tan ∅ = Q .....................Equation 5

In order to get a quadratic equation that can be easily solve.

800 = 2000 Q - 19.6 $Q^{2}$ + 19.6

Rearranging 19.6 $Q^{2}$ - 2000 Q + 780.4 = 0

$Q_{1}$ = 101.6291

$Q_{2}$ = 0.411

Inserting the value of Q Into Equation 5

tan ∅ = 101.63 or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

∅ = 89.44° ∅ = 22.37°

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2 years ago

Please anyone post any class 9 physics question

aalyn [17]

Answer:

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6 0

2 years ago

Joanne drives her car at a speed of 20 m/s. when she applied her breaks, a frictional force of 2000 N brought her car to a compl

Papessa [141]

Answer:

A) 1000 kg

Explanation:

vf = vi + at

0 = 20 + (a)(10)

a = -2.0 m/s^2

F = ma

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m = 1000 kg

8 0

9 months ago

Calculate the de Broglie wavelength of (a) a mass of 1.0 g traveling at 1.0 m s−1 , (b) the same, traveling at 1.00 × 105 km s−1

lesantik [10]

Answer:

a) $\lambda=6.63\times10^{-31}m$

b) $\lambda=6.63\times10^{-39}m$

c) $\lambda=9.97\times10^{-11}m$

d) $\lambda=4.03\times10^{-36}$ m

e)λ=∞

Explanation:

De Broglie discovered that an electron or other mass particles can have a wavelength associated, and that wavelength (λ) is:

$\lambda=\frac{h}{P}=\frac{h}{mv}$

with h the Plank's constant ( $6.63\times10^{-34}\frac{m^{2}kg}{s}$ ) and P the momentum of the object that is mass (m) times velocity (v).

a) $\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}kg*1.0)}$

$\lambda=6.63\times10^{-31}m$

b) $\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}*(1.00\times10^{8}))}$

$\lambda=6.63\times10^{-39}m$

c) $\lambda=\frac{6.63\times10^{-34}}{(6.65\times10^{-27}*1000)}$

$\lambda=9.97\times10^{-11}m$

d) $\lambda=\frac{6.63\times10^{-34}}{(74*2.22)}$

$\lambda=4.03\times10^{-36}$ m

e) $\lambda=\frac{6.63\times10^{-34}}{(74*0)}$

λ=∞

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3 years ago

A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If

Salsk061 [2.6K]

Complete question:

A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later?

Answer:

The current in the circuit 7 ms later is 0.2499 A

Explanation:

Given;

Ideal inductor, L = 45-mH

Resistor, R = 60-Ω

Ideal voltage supply, V = 15-V

Initial current at t = 0 seconds:

I₀ = V/R

I₀ = 15/60 = 0.25 A

Time constant, is given as:

T = L/R

T = (45 x 10⁻³) / (60)

T = 7.5 x 10⁻⁴ s

Change in current with respect to time, is given as;

$I(t) = I_o(1-e^{-\frac{t}{T}})$

Current in the circuit after 7 ms later:

t = 7 ms = 7 x 10⁻³ s

$I(t) = I_o(1-e^{-\frac{t}{T}})\\\\I =0.25(1-e^{-\frac{7*10^{-3}}{7.5*10^{-4}}})\\\\I = 0.25(0.9999)\\\\I = 0.2499 \ A$

Therefore, the current in the circuit 7 ms later is 0.2499 A

6 0

3 years ago