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3 years ago

9

A ball is kicked from a height of 20 meters above the ground. If the initial velocity is 10 m/s, how long is the ball in flight

before striking the ground?

1 answer:

nasty-shy [4]3 years ago

8 0

S=20 m
v=10 m/s
t=s/v
= 20/10
= 2 s.

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the length of iron rod at 100 C is 300.36 cm and at 159 C is 300.54 cm.Calculate its length at 0 c and coefficient of linear exp

Ugo [173]

Answer:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

Explanation:

From the question given above, the following data were obtained:

Length (L₁) at 100 °C = 300.36 cm

Temperature 1 (θ₁) = 100 °C

Length (L₂) at 159 °C = 300.54 cm

Temperature 2 (θ₂) = 159 °C

Length (L₀) at 0 °C =?

Coefficient of linear expansion (α) =?

L₁ = L₀ (1 + θ₁α)

300.36 = L₀ (1 + 100α) ....(1)

L₂ = L₀ (1 + θ₂α)

300.54 = L₀ (1 + 159α) ..... (2)

Divide equation (2) by (1)

300.54 / 300.36 = L₀ (1 + 159α) / L₀ (1 + 100α)

1.0006 = (1 + 159α) / (1 + 100α)

Cross multiply

1.0006 (1 + 100α) = (1 + 159α)

1.0006 + 100.06α = 1 + 159α

Collect like terms

1.0006 – 1 = 159α – 100.06α

0.0006 = 58.94α

Divide both side by 58.94

α = 0.0006 / 58.94

α = 1.02×10¯⁵ C¯¹

Substitute the value of α into anything of the equation to obtain L₀. Here we shall use equation (2).

300.54 = L₀ (1 + 159α)

α = 1.02×10¯⁵ C¯¹

300.54 = L₀ (1 + 159 ×1.02×10¯⁵)

300.54 = L₀ (1 + 0.0016218)

300.54 = L₀ (1.0016218)

Divide both side by 1.0016

L₀ = 300.54 / 1.0016

L₀ = 300.05 cm

Summary:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

6 0

3 years ago

a blackbody is radiating with a characteristic wavelength of 9 microns what is the blackbody temperature answer in kelvin

Daniel [21]

This question involves the concepts of Wein's displacement law and characteristic wavelength.

The blackbody temperature will be "3.22 x 10⁵ k".

<h3>WEIN'S DISPLACEMENT LAW</h3>

According to Wein's displacement law,

$\lambda_{max} T = c\\\\T=\frac{c}{\lambda_{max}}$

where,

$\lambda_{max}$ = characteristic wavelength = 9 μm = 9 x 10⁻⁹ m
T = temperature = ?
c = Wein's displacment constant = 2.897 x 10⁻³ m.k

Therefore,

$T=\frac{2.897\ x\ 10^{-3}\ m.k}{9\ x\ 10^{-9}\ m}$

T = 3.22 x 10⁵ k

Learn more about characteristic wavelength here:

brainly.com/question/14650107

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2 years ago

In a chemical equation, the chemicals that react are considered . In a chemical equation, the chemicals that are produced are co

vovangra [49]

Answer

Hi,

In a chemical equation, chemicals that react are the reactants, while chemicals that are produced are the products/by products. Both sides of the equation must be balanced.

Explanation

When writing a chemical equation, reactants reacts to produce products. For example in the equation for formation of water, hydrogen combines with oxygen as 2H₂ +O₂→2H₂O where the first part before the arrow represent the reactants and the next part after the arrow are the products. Reactants are on the left where as products are on the right.Coefficient 2, in this cases is used for balancing the equation.

Good luck!

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3 years ago

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During science class, Ava and Justin were discussing the rotation of the Sun and debating over which model would best illustrate

agasfer [191]

Answer:

it would b B

Explanation:

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3 years ago

A jet plane is launched from a catapult on an aircraft carrier. In 2.0 s it reaches a speed of 42 m/s at the end of the catapult

Dvinal [7]

Answer:

Explanation:

Given

time taken $t=2\ s$

Speed acquired in 2 sec $v=42\ m/s$

Here initial velocity is zero $u=0$

acceleration is the rate of change of velocity in a given time

$a=\frac{v-u}{t}$

$a=\frac{42-0}{2}=21\ m/s^2$

Distance travel in this time

$s=ut+0.5at^2$

where

s=displacement

u=initial velocity

a=acceleration

t=time

$s=0+\0.5\times 21\times (2)^2$

$s=42\ m$

so Jet Plane travels a distance of 42 m in 2 s

5 0

3 years ago