Jakie trudności spotykaja chrzescijan w głoszeniu ewangelii

A child riding a bicycle at 15 m/s accelerates at 3. 0 m/s/s for 4. 0 s. What is the child's speed at the end of this 4. 0 s int

mamaluj [8]

Answer:

The final speed of bicycle would be 27 m/s².

Step-by-step explanation:

Given :

$\small\red\bull$ Initial velocity (u) = 15 m/s
$\small\red\bull$ Acceleration (a) = 3.0 m/s²
$\small\red\bull$ Time = 4 seconds

To Find :

$\small\red\bull$ Final velocity (v)

Using Formula :

$\longrightarrow{\pmb{\sf{v = u + at}}}$

$\small\blue\star$ v = final velocity
$\small\blue\star$ u = initial velocity
$\small\blue\star$ a = acceleration
$\small\blue\star$ t = time taken

Solution :

Substituting the values in the formula to find the final velocity :

$\longrightarrow{\sf{v = u + at}}$

$\longrightarrow{\sf{v = 15 + (3.0 \times 4)}}$

$\longrightarrow{\sf{v = 15 + (3 \times 4)}}$

$\longrightarrow{\sf{v = 15 +(12)}}$

$\longrightarrow{\sf{v = 15 +12}}$

$\longrightarrow{\sf{v = 27 \: {m/s}^{2}}}$

$\star{\underline{\boxed{\red{\sf{v = 27 \: {m/s}^{2}}}}}}$

Hence, the final speed of bicycle would be 27 m/s².

$\rule{300}{1.5}$

3 0

3 years ago

How you can apply prudence and frugality on consuming water

34kurt

Answer:

The two main consumers of water are thermoelectric power and irrigation. So I'd say don't water your lawn as much and always turn your lights off when your not using them. And you could always take shorter showers.

Explanation:

3 0

2 years ago

How does the speaker's relationship to the tropical setting.

mart [117]

while the speaker appreciates the warmer climate at first

________________________________________________

8 0

3 years ago

A 100-n weight is hung at the middle of a 4-m long horizontal clothesline. If the middle of the clothesline sags by 1 m, what is

Elena L [17]

The tension on each segment of the clothesline is : 110 N

Given data :

mass of object = 100 n = 10 kg

Horizontal distance of clothesline = 4 m

middle of clothesline sag s by : 1m

<h3 /><h3>Determine the tension on each segment of clothesline</h3>

First step : calculate the horizontal angle made by the sagging

β = arctan ( 1 m / 2m ) ----- ( 1 )

= arctan ( 0.5 )

≈ 26.57°

Note : Tension in th y axis ( Ty ) = Tsinβ

Therefore :

Tension on each segment can be calculated using the formula below

2Tsinβ - mg = 0

solve for T

T = mg / 2sinβ

= ( 10 * 9.8 ) / 2 * sin 26.57°

= 98 / 0.89

= 110 N

Hence we can conclude that the tension on each segment of the clothesline is : 110 N

Learn more about Tension calculations : brainly.com/question/24994188

7 0

2 years ago

What mass of glucose (C6H12O6) should be dissolved in 12. 0 kg of water to obtain a solution with a freezing point of -5. 8 ∘C?.

bogdanovich [222]

The mass of glucose solute dissolved in the solution is 6.739 Kg.

Recall that;

ΔT = K m i

ΔT = Freezing point depression

K =Freezing point depression constant = 1.86°C/mol

m = molality of the solution

i = Van't Hoff factor = 1 (molecular solution)

We have to find the freezing point depression from;

Freezing point depression = Freezing point of pure solvent - Freezing point of solution

Freezing point of pure water = 0°C

Freezing point of solution = -5. 8 ∘C

Freezing point depression = 0°C - (-5. 8 ∘C) = 5. 8 ∘C

Now;

m = ΔT/K i

m = 5. 8 ∘C/ 1.86°C/mol × 1

m = 3.12 m

But molality = number of moles of solute/mass of solvent in Kg

Molar mass of solute = 180 g/mol

Let the mass of solute be m

3.12 = m/180/12

3.12 = m/180 × 12

m = 3.12 × 180 × 12

m = 6739 g or 6.739 Kg

Learn more: brainly.com/question/6249935

8 0

2 years ago

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Jakie trudności spotykaja chrzescijan w głoszeniu ewangelii​

Jakie trudności spotykaja chrzescijan w głoszeniu ewangelii