Noah stands 170 meters away from a steep canyon wall. He shouts and hears the echo of his

A ball is thrown with a velocity of 40 m/s at an angle of 30° above the horizontal and attains a certain range R. At what other

choli [55]

Answer:

This ball will attain the same range at 60°.

Explanation:

Projectile motion: when an object is thrown in such a way that it form an angle with horizon, the force act on it that is the acceleration due gravity. This type of motion is known as projectile motion.

Range: The horizontal distance is covered by an object.

Range $=\frac{u^2 sin2\theta}{g}$

u = initial velocity = 40 m/s

θ = 30°

g = gravity =9.8 m/s²

$Range=\frac{40^2\times sin(2\times 30^\circ)}{9.8}$

$=\frac{800\sqrt{3} }{9.8}$

Next,

Range $=\frac{800\sqrt{3} }{9.8}$ and u = 40 m/s

$sin2\theta =\frac{Range \times g}{u^2}$

$\Rightarrow sin2\theta =\frac{\frac{800\sqrt{3} }{9.8} \times 9.8}{40^2}$

$\Rightarrow sin 2\theta =\frac{\sqrt{3} }{2}$

$\Rightarrow sin 2\theta = sin 120^\circ$

$\Rightarrow \theta =120^\circ$

$\Rightarrow \theta =60^\circ$

This ball will attain the same range at 60°.

3 0

3 years ago

A copper sphere was moving at 25 m/s when it hit another object. this caused all of the ke to be converted into thermal energy f

muminat

ΔT= 81°C was the increase in temperature.

<h3>steps</h3>

As evidenced by energy conservation

To increase its temperature, all of its kinetic energy will be converted to thermal energy.

$\frac{1}{2}mv ^{2} = ms$ ΔT

then divide both sides by the object's mass.

$\frac{1}{2} v^{2} =s$ ΔT

therefore, a temperature shift is described as

ΔT= $V^{2} /2s$

ΔT= $25^{2}$ / 2 x 387

ΔT= 81°C

ΔT= 81°C was the increase in temperature.

learn more about increase in temperature here

<u>brainly.com/question/14319779</u>

#SPJ4

3 0

1 year ago

An object has a net charge of +5 Coulombs.

exis [7]

The net charge of the electron will be there because there is no exponents there

3 0

3 years ago

A box of spherical jawbreaker candies is 23 g and its volume is 32.3 cm3. If the average mass of a single jawbreaker is 0.94 g,

Alex777 [14]

The radius of each jawbreaker is approximately 0.68 cm.

<h3>Volume of a sphere;</h3>

v = 4 /3 πr³

where

r = radius

Therefore,

23 g = 32.3 cm³

0.94 g = ?

cross multiply

volume of a single jawbreaker = 32.3 × 0.94 / 23 = 30.362 / 23 = 1.32 cm³

Therefore,

volume of each jawbreaker = 4 /3 πr³

1.32 = 4 / 3 × 3.14 × r³

r³ = 1.32 /4.18666666667

r³ = 0.31533683707

r = ∛0.31533683707

r = 0.680651651 = 0.68

Therefore, the radius of each jawbreaker is approximately 0.68 cm.

learn more on radius here: brainly.com/question/19172427

5 0

3 years ago

On an aircraft carrier, a jet can be catapulted from 0 to 240

IRISSAK [1]

Answer:

m = 27752.7 [kg]

Explanation:

To solve this problem we must first use the following equation of kinematics.

$v_{f}=v_{o}+a*t$

where:

Vf = final velocity = 240 [km/h]

Vo = initial velocity = 0

a = acceleration [m/s²]

t = time = 2 [s]

But first we must convert the speed from kilometers per hour to meters per second.

$240[\frac{km}{h} ]*(\frac{1h}{3600s} )*(\frac{1000m}{1km} )=66.67[m/s]$

Now replacing:

$66.67 = 0 +a*2\\a = 33.33 [m/s^{2} ]$

Now using Newton's second law which is defined as the product of mass by acceleration we can determine the mass of the jet.

∑F = m*a

$925000=m*33.33\\m = 925000/33.33\\m = 27752.7 [kg]$

4 0

3 years ago