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Readme [11.4K]
2 years ago
5

A boat is traveling upstream at 14 km/h with respect to the water of a river. The water itself is flowing at 9 km/h with respect

to the ground. What is the velocity of the boat with respect to the ground
Physics
1 answer:
kirill [66]2 years ago
6 0

Answer:

16.64 km/h

Explanation:

Think of the x and y components as the a and b sides of a triangle and simply do the Pythagorean Theorem:

a^2+b^2=c^2

14^2+9^2=c^2

\sqrt{196+81}=c

\sqrt{277}=c

16.64 km/h

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Without the wheels, a bicycle frame has a mass of 8.29 kg. Each of the wheels can be roughly modeled as a uniform solid disk wit
trapecia [35]

Answer:

69.66 Joule

Explanation:

mass of bicycle frame, mf = 8.29 kg

mass of wheel, mw = 0.820 kg

radius, r = 0.343 m

velocity, v = 3.6 m/s

There are two wheels in the bicycle.

There are two types of kinetic energy of the system one is kinetic energy of rotation and another is rotational kinetic energy.

K = \frac{1}{2}m_{f}v^{2}+ 2\times \frac{1}{2}m_{w}v^{2}+ 2\times \frac{1}{2}I_{w}\omega^{2}

K = \frac{1}{2}m_{f}v^{2}+m_{w}v^{2}+ \frac{1}{2}\times m_{w}v^{2}

K = \frac{1}{2}m_{f}v^{2}+ \frac{3}{2}\times m_{w}v^{2}

K = \frac{1}{2}\times 8.29\times 3.6^{2}+ \frac{3}{2}\times 0.820\times 3.6^{2}

K = 69.66 J

3 0
3 years ago
A 60 g ball of clay is thrown horizontally at 40 m/s toward a 1.5 kg block sitting at rest on a frictionless surface. the clay h
Bingel [31]
The solution for this problem is:
Let u denote speed. 

Equating momentum before and after collision: 
= 0.060 * 40 = (1.5 + 0.060) u 
= 2.4 = 1.56 u
= 2.4 / 1.56 = 1.56 u / 1.56
= 1.6 m / s is the answer for this question. This is the speed after the collision.
7 0
2 years ago
Two charged objects are 1 meter apart. Calculate the magnitude of the electric force between them if the two charges are +1.0 μC
Solnce55 [7]

Answer:

0.00899 N

Explanation:

The magnitude of the electrostatic force between two charges is given by the equation:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the charges

r is the distance between the two charges

And the force is:

- Repulsive if the two charges have same sign

- Attractive if the two charges have opposite sign

In this problem we have:

q_1=1.0\mu C = 1.0\cdot 10^{6}C (charge of object 1)

q_2=1.0\mu C = 1.0\cdot 10^{6}C (charge of object 2)

r = 1 m (separation between the objects)

So, the electric force is

F=(8.99\cdot 10^9)\frac{(1.0\cdot 10^{-6})^2}{1^2}=0.00899 N

5 0
3 years ago
Everything is in the pic.
nikitadnepr [17]
The answer is A

Good luck!
3 0
3 years ago
A 6.89-nC charge is located 1.76 m from a 4.10-nC point charge. (a) Find the magnitude of the electrostatic force that one charg
iogann1982 [59]

Answer: a) 8.2 * 10^-8 N or 82 nN and b) is repulsive

Explanation: To solve this problem we have to use the Coulomb force for two point charged, it is given by:

F=\frac{k*q1*q2}{d^{2}}

Replacing the dat we obtain F=82 nN.

The force is repulsive because the points charged have the same sign.

5 0
3 years ago
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