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3 years ago

What is the relationship between the perimeter and the side length of anequilateral triangle ?

Physics

1 answer:

katovenus [111]3 years ago

3 0

Answer:

The length of the equilateral triangle's side is the length of the hypotenuse of the 30-60-90. Using this ratio, we find that the length of this triangle's hypotenuse is 4. Thus the perimeter of the equilateral triangle will be 4 multiplied by 3, which is 12.

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A sonar emits a sound signal of frequency 40000Hz, towards the bottom of the sea.

kap26 [50]

Answer:

0.000025s

Explanation:

Period it’s. : T(s)= 1/f(Hz)=1/40000Hz=0.000025s

8 0

3 years ago

How am I supposed to solve this?

RSB [31]

Answer:

4.02 km/hr

Explanation:

5 km/hr = 1.39 m/s

The swimmer's speed relative to the ground must have the same direction as line AC.

The vertical component of the velocity is:

uᵧ = us cos 45

uᵧ = √2/2 us

The horizontal component of the velocity is:

uₓ = 1.39 − us sin 45

uₓ = 1.39 − √2/2 us

Writing a proportion:

uₓ / uᵧ = 121 / 159

(1.39 − √2/2 us) / (√2/2 us) = 121 / 159

Cross multiply and solve:

159 (1.39 − √2/2 us) = 121 (√2/2 us)

220.8 − 79.5√2 us = 60.5√2 us

220.8 = 140√2 us

us = 1.115

The swimmer's speed is 1.115 m/s, or 4.02 km/hr.

7 0

3 years ago

Read 2 more answers

Big Ben, a large artifact in England, has a mass of 1x10^8 kilograms and the Empire State Building 1x10^9 kilograms. The distanc

TiliK225 [7]

Answer:

The force, exerted by Big Ben on the Empire State Building is 2.66972 × 10⁻⁷ N

Explanation:

The question relates to the force of gravity experienced between two bodies

The given parameters are;

The mass of Big Ben, M₁ = 1 × 10⁸ kg

The mass of the Empire State Building, M₂ = 1 × 10⁹ kg

The distance between the two Big Ben and the Empire State Building, r = 5,000,000 meters

By Newton's Law of gravitation, we have;

$F=G \times \dfrac{M_{1} \times M_{2}}{r^{2}}$

Where;

F = The force exerted by Big Ben on the Empire State Building and vice versa

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

M₁, M₂, and r are the given parameters

By plugging in the values of the parameters and the constant into the equation for Newton's Law of gravitation, we have;

$F=6.67430 \times 10^{-11} \times \dfrac{1 \times 10^8 \times 1 \times 10^9}{(5,000,000)^{2}} = 2.66972 \times 10^{-7}$

The force, 'F', exerted by Big Ben on the Empire State Building is F = 2.66972 × 10⁻⁷ N.

3 0

3 years ago

On your first trip to Planet X you happen to take along a 180 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You'r

lys-0071 [83]

Answer:

g_x = 3.0 m / s^2

Explanation:

Given:

- Change in length of spring [email protected] = 22.6 cm

- Time taken for 11 oscillations t = 19.0 s

Find:

- The value of gravitational free fall g_x at plant X:

Solution:

- We will assume a simple harmonic motion of the mass for which Time is:

T = 2*pi*sqrt(k / m ) ...... 1

- Sum of forces in vertical direction @equilibrium is zero:

F_net = k*x - m*g_x = 0

(k / m) = (g_x / x) .... 2

- substitute Eq 2 into Eq 1:

2*pi / T = sqrt ( g_x / x )

g_x = (2*pi / T )^2 * x

- Evaluate g_x:

g_x = (2*pi / (19 / 11) )^2 * 0.226

g_x = 3.0 m / s^2

3 0

3 years ago

In an elastic collision, a 580 kg bumper car collides directly from behind with a second, identical bumper car that is traveling

kozerog [31]

Answer: vl = 2.75 m/s vt = 1.5 m/s

Explanation:

If we assume that no external forces act during the collision, total momentum must be conserved.

If both cars are identical and also the drivers have the same mass, we can write the following:

m (vi1 + vi2) = m (vf1 + vf2) (1)

The sum of the initial speeds must be equal to the sum of the final ones.

If we are told that kinetic energy must be conserved also, simplifying, we can write:

vi1² + vi2² = vf1² + vf2² (2)

The only condition that satisfies (1) and (2) simultaneously is the one in which both masses exchange speeds, so we can write:

vf1 = vi2 and vf2 = vi1

If we call v1 to the speed of the leading car, and v2 to the trailing one, we can finally put the following:

vf1 = 2.75 m/s vf2 = 1.5 m/s

8 0

3 years ago