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lianna
1 month ago
12

An Object moving at a velocity of 30 m/s slows to a stop in 7 seconds. What was its acceleration

Physics
1 answer:
inna [3.1K]1 month ago
7 0

Answer:

The result is 4.29 m/s²

Explanation:

The acceleration of an object can be calculated using its velocity and the duration of time using a specific formula.

acceleration = \frac{velocity}{time} \\

According to the problem:

velocity = 30 m/s

time = 7 s

Thus, we find the final answer:

4.29 m/s²

I hope this information is useful to you.

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9. A 2 liter bottle of Coke weighs about 2 kilograms. If that bottle were combined with an equal amount of anti-Coke, how many J
Maru [3345]

Answer:

The energy expected to be released is calculated to be 4182 Joules.

Explanation:

The total mass of coke is 2 kg, which is equivalent to 2000 g

1 calorie per gram corresponds to 4.184 Joules of energy

4.184 J/gC * 2000g results in 8368 J

1 food calorie approximates to 4186 J

By subtracting, we find 8368 - 4186

Hence, the total energy that will be released amounts to 4182 Joules.

3 0
1 month ago
A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per
Softa [3030]

Result: -50.005 kJ

Details:

Provided Data

mass of the system = 10 kg

work done = 0.147 kJ/kg

Elevation change (\Delta h)=-50 m

initial speed (v_1)=15 m/s

Final Speed (v_2)=30 m/s

Specific internal Energy (\Delta U)=-5 kJ/kg

according to the first Law of thermodynamics

Q-W=\Delta H

\Delta H=\Delta KE+ \Delta PE +\Delta U

where KE represents kinetic energy

PE indicates potential energy

U denotes internal Energy

\Delta H=\frac{m(v_2^2-v_1^2)}{2}+mg(\Delta h)+\Delta U

Q=W+\Delta KE+ \Delta PE +\Delta U

Q=0.147\times 10+\frac{10\cdot (30^2-15^2)}{2}+10\cdot 9.81\cdot (-5)-5\times 10

Q = 1.47 + 3.375 - 4.850 - 50

Q = -50.005 kJ

5 0
2 months ago
Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of -2.0 µC; sphere B carries a charge of -6.0 µC;
inna [3103]
None of the provided options is correct. After contact, A becomes -4 µC, B remains 0 µC, and C ends with +4.0 µC. When spheres A and B touch, charges will redistribute to establish balance, resulting in A = -4 µC, B = -4 µC, C = +4.0 µC. After C and B are touched, both positive and negative charges neutralize each other, leaving A at -4 µC, B at 0 µC, and C at 0 µC.
5 0
1 month ago
In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we
Yuliya22 [3333]

Answer:

The typical weight of a human heart is approximately 0.93 lbs.

Explanation:

Based on this,

the heart's weight constitutes about 0.5% of total body mass.

Total human weight = 185 lbs

Let the entire body weight be represented as w and the heart's weight as w_{h}.

We aim to determine the heart's weight for a human

Using the provided information

w_{h}=0.5\times w

Where, h = heart weight

w = human weight

w_{h}=\dfrac{0.5}{100}\times 185

w_{h}=0.93\ lbs

The final weight of a human heart is 0.93 lbs.

8 0
1 month ago
A transverse wave on a string has an amplitude A. A tiny spot on the string is colored red. As one cycle of the wave passes by,
Sav [3153]
Since it's classified as a transverse wave, the particle on the string moves horizontally as the wave progresses, without actual forward or backward travel. Consequently, the red dot shifts 'A' to the left, returns 'A' to the center, moves 'A' to the right, and goes back 'A' to the center once again. Thus, the red dot collectively travels a distance totaling 4A.
6 0
1 month ago
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