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sergeinik [125]
3 years ago
14

A swimmer swims 40 miles in 80 seconds. What is the swimmer’s average speed?

Physics
1 answer:
Mamont248 [21]3 years ago
6 0

Answer:2

Explanation:

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1. An object of mass 300 kg is observed to
vekshin1

Answer:

The force required will be "300 N".

Explanation:

The given values are:

Mass of object,

m = 300 kg

Acceleration,

a = 1 m/s

Now,

The force will be:

⇒  F=m\times a

On substituting the values, we get

⇒      = 300\times 1

⇒      =300 \ N

4 0
3 years ago
A 5.0 kg box is sliding across a waxed floor by the application of a 15 N east force. If the force of friction is 2.5 N west, wh
coldgirl [10]

1) 12.5 N east

There are two forces acting on the box along the horizontal direction:

- The applied force of 15 N east, we indicate it with F

- The force of friction of 2.5 N west, we indicate it with F_f

Taking east as positive direction, we can write the two forces has

F=+15 N\\F_f = -2.5 N

Therefore, the net force on the box will be:

F_{net} = F + F_f = 15 + (-2.5) = +12.5 N

And the positive sign means the direction is east.

2) 2.5 m/s^2

We can solve this part by using Newton's second law:

F_{net}=ma

where

F_{net} is the net force on the box

m is its mass

a is the acceleration

For the box in this problem,

F_{net} = 12.5 m/s^2 (east)

m = 5.0 kg

Solving for a, we find the acceleration:

a=\frac{F_{net}}{m}=\frac{12.5}{5.0}=2.5 m/s^2

And the direction is the same as the net force (east)

6 0
3 years ago
Bernoulli's Principle is a consequence of the conservation of
Ilia_Sergeevich [38]
The answer for this would be B) Energy
4 0
2 years ago
What divides the northern hemisphere from the southern hemisphere
asambeis [7]
The equator - hope I helped you :)
7 0
3 years ago
A radio antenna broadcasts a 1.0 MHz radio wave with 30 kW of power. Assume that the radiation is emitted uniformly in all direc
kozerog [31]

Answer:

a) intensity  = 2.19 \times  10^{-6} W/m^2

b) E = 0.0406 V/m

Explanation:

given data:

Power =  30 kW

R = 33×10^3 M

1) Intensity is given as

Intensity I = \frac{P}{4\times \pi \times r^2}

= \frac{30 \times 10^3}{4\times 3.14 \times (33\times10^3)^2}

= 2.19 \times  10^{-6} W/m^2

2) electric field is given as

I = 1/2 \times \epsilon \times c \times E^2

Solving for E

E = \sqrt{\frac{2I}{\epsilon c}}

E  =\sqrt{\frac{2\times 2.19 \times 10^{-6}}{8.85\times 10^{-12}  3\times 10^8}}

E = 0.0406 V/m

6 0
3 years ago
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