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beks73 [17]
3 years ago
5

After traveling for 6.0 seconds, a runner reaches 10m/s. What is the runner's acceleration?

Physics
1 answer:
luda_lava [24]3 years ago
4 0

After traveling for 6.0 seconds, a runner reaches 10m/s. What is the runner's acceleration? Answer: 1.67 m/s2

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25.16 what is the reaction product of acetic acid and ethylamine at room temperature?
Ne4ueva [31]

The ammonium salt of acetic acid is the reaction product of acetic acid and ethylamine at room temperature

<h3 /><h3>What is acetic acid ?</h3>

Acetic acid is a monofunctional carboxylic acid containing two carbon atoms. It acts as a protein solvent,  food acidity regulator, antibacterial food preservative. It is a conjugate acid of an acetate.

Acetic acid is used in the production  of acetic anhydride, cellulose acetate, vinyl acetate monomer, acetic ester, chloroacetic acid, plastics, dyes, insecticides, photographic chemicals, and rubber. Other commercial uses include the production of vitamins, antibiotics, hormones,  organic chemicals, and as a food additive. Typical concentrations of acetic acid found naturally in foods are 700 to 1200 milligrams/kg (mg/kg) in wine, up to 860 mg/kg in aged cheeses, and 2.8 mg/kg in aged cheeses. fresh orange juice.

 

learn more about acetic acid, visit;

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7 0
1 year ago
How was fibre optics made​
EastWind [94]
The first working fiber-optic data transmission system was demonstrated by German physicist Manfred Börner at Telefunken Research Labs in Ulm in 1965, which was followed by the first patent application for this technology in 1966. In 1968, NASA used fiber optics in the television cameras that were sent to the moon.
3 0
3 years ago
The car has a mass of 1.6 Mg and center of mass at G. If the coefficient of static friction between the shoulder of the road and
Rufina [12.5K]

Answer:

\theta = 22.29

Explanation:

Taking summation of force at perpendicular  to the plane

\sum F_p = 0

2N_A +2N_B -mgcos\theta = 0

N_A +N_B - mgcos\theta = 0

N_A +N_B = mgcos\theta

Taking summation along the plane, therefore we have

\sum Fa = 0

f_A +f_B -mgsin\theta = 0  

\mu N_A+\mu N_B - mgsin\theta = 0

\mu(N_A +N_B) = mgcos\theta

from equation 1 and 2 we have

\mu =\frac{sin\theta}{cos\theta}

\mu = 0.41

\mu = tan\theta

\theta = tan^{-}\mu

\theta = 22.29

6 0
4 years ago
A 4.00 kg block is suspended from a spring with k 500 N/m. A 50.0 g bullet is fired into the block from directly below with a sp
Yanka [14]

Answer:

a. A = 0.1656 m

b. % E = 1.219

Explanation:

Given

mB = 4.0 kg , mb = 50.0 g = 0.05 kg , u₁ = 150  m/s , k = 500 N / m

a.

To find the amplitude of the resulting SHM using conserver energy

ΔKe + ΔUg + ΔUs = 0

¹/₂ * m * v²  -  ¹/₂ * k * A² = 0

A = √ mB * vₓ² / k

vₓ = mb * u₁ / mb + mB

vₓ = 0.05 kg * 150 m / s / [0.050 + 4.0 ] kg = 1.8518

A = √ 4.0 kg * (1.852 m/s)²   /   (500 N / m)

A = 0.1656 m

b.

The percentage of kinetic energy

%E = Es / Ek

Es = ¹/₂ * k * A² = 500 N / m * 0.1656²m = 13.72 N*0.5

Ek = ¹/₂ * mb * v² = 0.05 kg * 150² m/s = 1125 N

% E = 13.72 / 1125 = 0.01219 *100

% E = 1.219

6 0
3 years ago
Heat is transferred at a rate of 2 kW from a hot reservoir at 800 K to a cold reservoir at 300 K. Calculate the rate at which th
vekshin1

Answer:

0.00417 kW/K or 4.17 W/K

Second law is satisfied.

Explanation:

Parameters given:

Rate of heat transfer, Q = 2kW

Temperature of hot reservoir, Th = 800K

Temperature of cold reservoir, Tc = 300K

The rate of entropy change is given as:

ΔS = Q * [(1/Tc) - (1/Th)]

ΔS = 2 * (1/300 - 1/800)

ΔS = 2 * 0.002085

ΔS = 0.00417 kW/K or 4.17 W/K

Since ΔS is greater than 0, te the second law of thermodynamics is satisfied.

6 0
3 years ago
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