All categories

3 years ago

A bike travels at a constant speed for 4.00 m/s for 5.00 seconds. How far does it go

Physics

2 answers:

Umnica [9.8K]3 years ago

7 0

Answer:

20 m

Explanation:

4.00 m/s x 5 seconds

4 m/s x 5 s = 20m

Bumek [7]3 years ago

4 0

Answer:20.00

Explanation:

5×4=20

Bring down the zeros and you have 20.00

You might be interested in

4. A 15 V battery is connected to three capacitors in series. The capacitors have the following

Anastaziya [24]

Vp = Vs 14.o V battery 3.96 ûF

5 0

3 years ago

radio signals emitted from and received by an airplane have her fruit frequency of 3.00 * 10 to the 12th Hertz and travel at the

Elanso [62]

here we know that the speed of the signal is same as the speed of light

so here we will have

$v = 3 \times 10^8 m/s$

the altitude of the airplane is given as

$h = 1 \times 10^4 m$

now we know that time taken by the signal to reach the control tower is given as

$t = \frac{d}{v}$

now it is given as

$t = \frac{1 \times 10^4 }{3 \times 10^8}$

$t = 3.33 \times 10^{-5} s$

so above is the time taken by the signal to reach the control tower

3 0

3 years ago

What is the wavelength of the carrier wave of a campus radio station, broadcasting at a frequency of 97.2 MHz (million cycles pe

andrey2020 [161]

Answer:

Wavelength is 3.08 meters.

Explanation:

It is given that,

Frequency, $f=97.2\ MHz=97.2\times 10^6\ Hz$

We need to find the wavelength of the carrier wave of a campus radio station. It can be calculated by using following relation as :

$c=f\times \lambda$

$\lambda=\dfrac{c}{f}$

$\lambda=\dfrac{3\times 10^8\ m/s}{97.2\times 10^6\ Hz}$

$\lambda=3.08\ m$

So, the wavelength of the carrier wave is 3.08 meters. Hence, this is the required solution.

8 0

3 years ago

Can a blackhole be destroyed?

lilavasa [31]

Nope a black hole cannot be destroyed it can however destroy everything else

5 0

3 years ago

Read 2 more answers

Determine the magnitude of the force F1F1 component acting along the uu axis. Express your answer to three significant figures a

dybincka [34]

The first image below shows force F1 and the axes.

Answer: $(F_{1})_{u}=$ 3.62 kN

Explanation: The second figure below express the parallelogram method to calculate the u component of force F1.

The Parallelogram Method is a method to determine resultant force and is applied as described in the question above.

With the three components, $F_{1},(F_{1})_{u}$ and $(F_{1})_{v}$ and angles, it can be used the Law of Sines, which states:

$\frac{a}{sin\alpha} =\frac{b}{sin\beta} =\frac{c}{sin\theta}$

i.e., there is a relation of proportionality between an angle and its opposite side.

For the triangle below:

$\frac{u}{sin30} =\frac{F_{1}}{sin105}$

$u=F_{1}\frac{sin(30)}{sin(105)}$

$u=7\frac{0.5}{0.966}$

u = 3.62

The magnitude of the component acting along the u-axis is 3.62kN.

5 0

3 years ago