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asambeis [7]
3 years ago
15

A bird flies with a constant velocity of 100m/s for 10 minutes. calculate the magnitude of the displacement.​

Physics
2 answers:
Ostrovityanka [42]3 years ago
4 0

Answer:

s = ut

= 100 m/s * 10*60 s

= 60000m

= 60km

Sedbober [7]3 years ago
3 0

Answer:

60 km

Explanation:

velocity (v) = 100m/s

t = 10 minutes = 10(60) = 600 seconds

displacement = velocity × time

s = v × t

s = 100 × 600

s = 60,000m

s = 60,000/1000 km

s = 60km

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In certain cases, using both the momentum principle and energy principle to analyze a system is useful, as they each can reveal
kramer

Explanation:

The gravitational force equation is the following:

F_G = G * \frac{m_1 m_2}{r^2} \\

Where:

G = Gravitational constant = 6.67408 * 10^{-11} m^3 kg^{-1} s^{-2}

m1 & m2 = the mass of two related objects

r = distance between the two related objects

The problem gives you everything you need to plug into the formula, except for the gravitational constant. Let me know if you need further clarification.

8 0
3 years ago
If a cross country runner covers a distance of 347 meters in 134 seconds what is her average speed?
a_sh-v [17]
347÷134=2.589552239 meters per second

2.589552239×60= 155.3731343 meters per hour

155.3731343 meters per hour= 0.096544389701642 miles per hour

hopefully this was right.
6 0
4 years ago
Consider the position vs. time graph below for a woman's movement in a hallway. What is the woman's velocity from 4 to 5 s?
Ksenya-84 [330]

Answer:

The answer is "6\  \frac{m}{s}"

Explanation:

The formula for velocity:

\to \overline{v}={\frac{\Delta x}{\Delta t}}

      =\frac{6}{1}\\\\=6\  \frac{m}{s}

7 0
3 years ago
Which of the following describes a referee's job?
Serhud [2]

Answer:

C. Supervising the game to make sure teams are playing fairly

5 0
4 years ago
A positive charge +q1 is located to the left of a negative charge -q2. On a line passing through the two charges, there are two
AfilCa [17]

Answer:

please the answer below

Explanation:

(a) If we assume that our origin of coordinates is at the position of charge q1, we have that the potential in both points is

V_1=k\frac{q_1}{r-1.0}-k\frac{q_2}{1.0}=0\\\\V_2=k\frac{q_1}{r+5.2}-k\frac{q_2}{5.2}=0\\\\

k=8.89*10^9

For both cases we have

k\frac{q_1}{r-1.0}=k\frac{q_2}{1.0}\\\\q_1(1.0)=q_2(r-1.0)\\\\r=\frac{q_1+q_2}{q_2}\\\\k\frac{q_1}{r+5.2}=k\frac{q_2}{5.2}\\\\q_1(5.2)=q_2(r+5.2)\\\\r=\frac{5.2q_1-5.2q_2}{q_2}

(b) by replacing this values of r in the expression for V we obtain

k\frac{q_1}{\frac{5.2(q_1-q_2)}{q_2}+5.2}=k\frac{q_2}{5.2}\\\\\frac{q_1}{q_2}=\frac{(q_1-q_2)}{q_2}-1.0=\frac{q_1-q_2-q_2}{q_2}=\frac{q_1-2q_2}{q_2}

hope this helps!!

3 0
3 years ago
Read 2 more answers
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