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2 years ago

6

A(n) _______ trial helps a scientist by strengthening the support of the hypothesis or strengthening the rejection of the hypoth

esis.

1 answer:

andreyandreev [35.5K]2 years ago

4 0

I believe the answer is "experimental"

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A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.

ivann1987 [24]

Answer:

v₂ = 176.24 m/s

Explanation:

given,

angle of projectile = 45°

speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

$H_{max}= \dfrac{v^2 sin^2(\theta)}{g}$

now,

$H_{max}= \dfrac{v_1^2 sin^2(\theta_1)}{g}$

$H_{max}= \dfrac{v_2^2 sin^2(\theta_2)}{g}$

now, equating both the equations

$\dfrac{v_2^2}{v_1^2}=\dfrac{sin^2(\theta_1)}{sin^2(\theta_2)}$

$\dfrac{v_2^2}{150^2}=\dfrac{sin^2(45^0)}{sin^2(37^0)}$

v₂² = 31061.79

v₂ = 176.24 m/s

velocity of projectile would be equal to v₂ = 176.24 m/s

8 0

2 years ago

PLEASE HELP FAST!!!

ELEN [110]

Answer:

1.It improved by using the Mars survey probe, that took a portrait of the planet that helped us see the climate of Mars.

Explanation:

I watched the video.

7 0

2 years ago

How do you think that changing the mass of the pendulum bob will affect the period of the pendulum swing?

myrzilka [38]

(Mass does not affect the pendulum's swing. The longer the length of string, the farther the pendulum falls; and therefore, the longer the period, or back and forth swing of the pendulum. The greater the amplitude, or angle, the farther the pendulum falls; and therefore, the longer the period.)

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2 years ago

Please help, it's an exam ://

LenaWriter [7]

Answer:

Whats the question

Explanation:

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2 years ago

Two long parallel wires are separated by forty centimeters and carry oppositely-directed currents of ten amperes. Find the magni

Softa [21]

Answer:

1.04μT

Explanation:

Due to both wires have opposite currents, the magnitude of the total magnetic field is given by

$B_T=\frac{\mu_o I}{2 \pi r_1}-\frac{\mu_o I}{2 \pi r_2}$

I: electric current = 10A

mu_o: magnetic permeability of vacuum = 4pi*10^{-7} N/A^2

r1: distance from wire 1 to the point in which B is measured.

r2: distance from wire 2.

The distance between wires is 40cm = 0.4m. Hence, r1=0.2m r2=0.6m

By replacing in the formula you obtain:

$B_T=\frac{(4\pi *10^{-7}N/A^2)(10A)}{2\pi}(\frac{1}{0.4m}-\frac{1}{0.6m})=1.04*10^{-6}T =1.04\mu T$

hence, the magnitude of the magnetic field is 1.04μT

4 0

3 years ago