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mamaluj [8]
3 years ago
6

Alvin paddled for 44 hours with a 33​-km/h current to reach a campsite. The return trip against the same current took 77 hours.

Find the speed of the boat in still water.
Physics
1 answer:
sveta [45]3 years ago
8 0

Answer:

121km/h

Explanation:

Let speed of boat=xkm/h

Speed of water flow=33 km/h

Speed upstream=(x-33)km/h

Speed downstream=(x+33)km/h

Time=44 hrs

Time =77 hrs

We have to find the speed of boat.

Distance=Speed\times time

Using the formula

Upstream distance=Downstream distance

(x-33)\times 77=(x+33)\times 44

77x-2541=44x+1452

77x-44x=1452+2541

33x=3993

x=\frac{3993}{33}=121

Hence, the speed of boat=121km/h

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A bicyclist covers the first leg of a journey that is d1meters in t1seconds at a speed of v1m/s and the second leg of d2meters i
Murljashka [212]

According to motion in straight line  t1≠t2


A biker travels d1 meters in t1 seconds at v1 m/s for the first leg and d2 meters in t2 seconds at v2 m/s for the second leg. It's possible that t1t2 if his average speed is equal to the average of v1 and v2.
An object is said to be in motion if its position in relation to its surroundings changes over time. It is a shift in an object's position over time. The only type of motion that exists is motion in a straight line.


A reference system is constantly used to describe a particle's motion. An arbitrary origin point is used to create a reference system, and a coordinate system is imagined to be connected to it. The reference system for a specific problem is the coordinate system that has been selected for it. For the majority of the problems, we typically select an earth-based coordinate system as the reference system.


To learn more about Motion in straight lines please visit -brainly.com/question/17675825
#SPJ1

4 0
2 years ago
A sports car is advertised to be able to stop in a distance of 50.0 m from a speed is 99.0km/h. What is its acceleration in m/s2
Shtirlitz [24]
99.0km/h =27.5m/s (this is the initial speed)
The final speed is zero
The distance is 50.0m
Therefore you use the formula:
vfinal²=vinitial²+2ad
a=(vfinal²-vinitial²)/2d
 = (0²-27.5²)/(2x50.0)
 =-7.5625 or in correct sigdigs -7.56m/s²
Hope this helps!
4 0
3 years ago
A battery is replaced with one of lower emf. State and explain how the resistance of the lamps would have to change in order to
kow [346]

The brightness of the lamp is proportional to the current flowing through the lamp: the larger the current, the brighter the lamp.

The current flowing through the lamp is given by Ohm's law:

I=\frac{V}{R}

where

V is the potential difference across the lamp, which is equal to the emf of the battery, and R is the resistance of the lamp.

The problem says that the battery is replaced with one with lower emf. Looking at the formula, this means that V decreases: if we want to keep the same brightness, we need to keep I constant, therefore we need to decrease R, the resistance of the lamp.

3 0
3 years ago
A 0.0663 kg ingot of metal is heated to 241◦C
Westkost [7]

Answer:280.216j/kg°C

Explanation:

Mass of metal=0.0663kg

mass of water=0.395kg

Final temperature=27.4°C

Temperature of metal=241°C

Temperature of water=25°C

specific heat capacity of water=4186j/kg°C

0.0663xax(241-27.4)=0.395x4186x(27.4-25)

0.0663xax213.6=0.395x4186x2.4

14.16168a=3968.328

a=3968.328 ➗ 14.16168

a=280.216j/kg°C

4 0
3 years ago
A wheel has a radius of r = 2.0 m and it rolls down a smooth incline. The height of the incline is h = 8.0 m . What is the angul
zepelin [54]

The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

i.e.

P.E = K.E + R.K.E

\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}

\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}

\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 +  \omega^2}

\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2}  }

\mathbf{\omega^2=\dfrac{39.24 }{2}}

\mathbf{\omega=\sqrt{19.62 } \ rad/sec}

\mathbf{\omega=4.429 \ rad/sec}

Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

Learn more about angular velocity here:

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4 0
2 years ago
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