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leva [86]
3 years ago
9

A bug is moving along a meter stick in the negative direction at a constant speed of 0.85cm/s. After 42 seconds have passed the

bug is located at 27.0 cm. What was the bug’s starting position.
Physics
1 answer:
Agata [3.3K]3 years ago
3 0

Given that,

bug speed, v= 0.85 m/s

time, t =42 s

Final position of bug on meter stick was 27 cm

Starting position of bug on meter stick = ?

Since we know that,

s = vt

s= 0.85*42 = 35.7 cm

this is the distance covered by bug in the given time and velocity.

since the bug is moving in negative direction, starting point will be:

27.0 cm+ 35.7 cm = 62.7 cm

The bugs starting position on meter stick was 62.7 cm.

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The water in a river flows uniformly at a constant speed of 2.50 m/s between parallel banks 80.0 m apart. You are to deliver a p
NISA [10]

Answer:

a)  The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) 133.33 m

c) 53.13°

d) 106.67 m

Explanation:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) velocity = distance * time

Let the velocity of the swimmer be v_{s} = 1.5 m/s

The separation of the two sides of the river, d = 80 m

The time taken by the swimmer to get to the other end of the river bank,

t = \frac{d}{v_{s} }

t = 80/1.5

t = 53.33 s

The swimmer will be carried downstream by the river through a distance, s

Let the velocity of the river be v_{r} = 2.5 m/s

S = v_{r} t

S = 53.33 * 2.5

S = 133.33 m

c) To minimize the distance traveled by the swimmer, his resultant velocity must be perpendicular to the velocity of the swimmer relative to water

That is ,

cos \theta = \frac{v_{s} }{v_{r} } \\cos \theta = 1.5/2.5\\cos \theta = 0.6\\\theta = cos^{-1} 0.6\\\theta = 53.13^{0}

d) Downstream velocity of the swimmer, v_{y} = v_{s} sin \theta\\

v_{y} = 1.5 sin 53.13\\v_{y} = 1.2 m/s

The vertical displacement is given by, y = v_{y} t

80 = 1.2 t

t = 80/1.2

t = 66.67 s

the horizontal speed,

v_{x} = 2.5 - 1.5cos53.13\\v_{x} = 1.6 m/s

The downstream horizontal distance of the swimmer, x = v_{x} t

x = 1.6 * 66.67

x = 106.67 m

7 0
3 years ago
You push a table 8 meters for 16 minuutes and do 6720 j of work how much power do ypu use
Yuri [45]
P=W/t

P=Power
W=Work
t=Time

Convert 16 minutes in seconds:
16 mins = 960 secs

P=6720/960=7.23 W [Watt]
3 0
3 years ago
Radon-220 is a radioactive gas. It escapes from underground rocks and causes a large part of the natural background radiation in
Sergeu [11.5K]

Answer: 134 neutrons are there in the nucleus of Radon-220

Explanation:

Atomic number : It is defined as the number of electrons or number of protons present in a neutral atom.

Thus, number of protons = atomic number = 86

Mass number is the number of the entities present in the nucleus which is the equal to the sum of the number of protons and electrons.

Mass number = Number of protons + Number of neutrons

220 = 86 + Number of neutrons

Number of neutrons = 134

5 0
3 years ago
NaOH + FeCl3* Na Cl + Fe 10H)3<br> balanced
zvonat [6]
3NaOH + FeCl3 → 3NaCl + Fe(OH)3
8 0
3 years ago
If two particles have equal kinetic energies, are their momenta necessarily equal? explain.
Mandarinka [93]

Answer:

No the given statement is not necessarily true.

Explanation:

We know that the kinetic energy of a particle of mass 'm' moving with velocity 'v' is given by

K.E=\frac{1}{2}mv^{2}

Similarly the momentum is given by m\times v

For 2 particles with masses m_{1},m_{2}and moving with velocities v_{1},v_{2} respectively the respective kinetic energies is given by

K.E_{1}=\frac{1}{2}m_{1}v_{1}^{2}

K.E_{2}=\frac{1}{2}m_{2}v_{2}^{2}

Similarly For 2 particles with masses m_{1},m_{2}and moving with velocities v_{1},v_{2} respectively the respective momenta are given by

p_{1}=m_{1}\times v_{1}

p_{2}=m_{2}\times v_{2}

Now since it is given that the two kinetic energies are equal thus we have

\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{2}v_{2}^{2}\\\\(m_{1}v_{1})\times v_{1}=(m_{2}v_{2})\times v_{2}\\\\p_{1}\times v_{1}=p_{2}\times v_{2}\\\\\therefore \frac{p_{1}}{p_{2}}=\frac{v_{2}}{v_{1}}............(i)

Thus we infer that the moumenta are not equal since the ratio on right of 'i' is not 1 , and can be 1 only if the velocities of the 2 particles are equal which becomes a special case and not a general case.

5 0
3 years ago
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