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Dafna1 [17]
3 years ago
7

2. A boy kicks the football of mass 500 g which is initially at rest. It starts moving with a

Physics
2 answers:
Zarrin [17]3 years ago
8 0

Answer:

mass 500g is converted into kg

500/1000=0.5kg

same 1/2=0.5

Explanation:

nikdorinn [45]3 years ago
4 0

Answer:

Below given

Explanation:

constant velocity of 5 m/s. Calculate:

(a) the kinetic energy of the boy

(b) the work done by the boy

[Ans.(a) 6.25 ), (b) 6.25]] can some on show me how the answer

came

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When determining whether a chemical reaction has taken place, you observe and look for several indicators. Which would be consid
Inga [223]
The answer is B.

A cannot be the answer is melting is a physical change. No chemical reaction took place.

B is the answer as it is a EXOTHERMIC REACTION so heat will be given off.

C cannot be the answer as dissolving is basically atoms becoming ions, not a chemical reaction whereby a reactant reacts with another reactant to form a product.

D cannot be the answer. Same reason as for why A is not the answer.

Cheers.
6 0
3 years ago
A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the
lana [24]
(a) Let's convert the final speed of the car in m/s:
v_f = 61 km/h = 16.9 m/s
The kinetic energy of the car at t=19 s is
K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
P= \frac{W}{\Delta t}
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W

(c) The instantaneous power is given by
P_i = Fv_f
where F is the force exerted by the engine, equal to F=ma.

So we need to find the acceleration first:
a= \frac{v_f-v_i}{\Delta t}=  \frac{16.9 m/s}{19 s}=0.89 m/s^2
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
P_i = Fv=(ma)v=(1400 kg)(0.89 m/s^2)(16.9 m/s)=2.11 \cdot 10^4 W
5 0
3 years ago
Traveling waves are generated on a string fixed at both ends. The string has a length L, a linear mass density m, and a tension
bagirrra123 [75]

Answer: d. I or II

Explanation: A traveling wave has speed that depends on characteristics of a medium. Characteristics like linear density (μ), which is defined as mass per length.

Tension or Force (F_{T}) is also related to the speed of a moving wave.

The relationship between tension and linear density and speed is ginve by the formula:

|v|=\sqrt{\frac{F_{T}}{\mu} }

So, for the traveling waves generated on a string fixed at both ends described above, ways to increase wave speed would be:

1) Increase Tension and maintaining mass and length constant;

2) Longer string will decrease linear density, which will increase wave speed, due to their inversely proportional relationship;

Then, ways to increase the wave speed is

I. Using the same string but increasing tension

II. Using a longer string with the same μ and T.

8 0
3 years ago
Which is a correct step in a scientific experiment involving acids and bases?
Gre4nikov [31]
The answer, using an indicator to measure the hydrogen ion concentration of a solution, is correct
4 0
3 years ago
Read 2 more answers
Four capacitors with capacitance 3.0 pF, 2.0 pF, 5.0 pF and X pF are connected in series to each other. If the equivalent capaci
bagirrra123 [75]

Answer:

<h2>A. 6pF</h2>

Explanation:

If unknown capacitance C1, C2, C3 and C4 are connected in series to one another, their equivalent capacitance of the circuit will be expressed as shown

\frac{1}{C_t} = \frac{1}{C_1} +\frac{1}{C_2} +\frac{1}{C_3} +\frac{1}{C_4} \\

Given the capacitance's 3.0 pF, 2.0 pF, 5.0 pF and X pF connected in series to each other. If the equivalent capacitance of the circuit is 0.83 pF, then to get X, we will apply the formula above;

\frac{1}{0.83} = \frac{1}{3.0} +\frac{1}{2.0} +\frac{1}{5.0} +\frac{1}{C_4} \\\\\\1.205 = 0.333+0.5+0.2+\frac{1}{C_4} \\\\1.205 = 1.033 + \frac{1}{C_4} \\\\\frac{1}{C_4}  = 1.205-1.033\\\\\frac{1}{C_4}  = 0.172\\\\C_4 = \frac{1}{0.172}\\ \\C_4 = 5.8pF\\\\

C₄ ≈ 6pF

Hence the value of the X capacitor is approximately 6pF

8 0
3 years ago
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