All categories

3 years ago

Which formula can be used to calculate average speed (v) for steady motions?

Physics

1 answer:

Liono4ka [1.6K]3 years ago

3 0

I’m pretty sure it’s speed=distance/time

You might be interested in

a charge of 7.2x10^-5C is placed in an Electric field with a strength of 4.8x10^5N/C. if the Electric potential energy of the ch

Alecsey [184]

Answer:

$d= 2.2 m$

Explanation:

The data given in the question is

Charge : q = $7.2 \times 10^{-5} C$

Electric field Strength : E = $4.8 \times 10^{5} N/C$

Electrical potential Energy : U = $75J$

Let, the distance be "d"

So, the formula for Electrical potential energy is

$U= q \times E \times d$

Simplified formula for distance become

$d=\frac{U}{q \times E}$

Now , insert the value

$d=\frac{75 J}{7.2 \times 10^{-5}C \times 4.8 \times 10^5 N/C}$

or,

$d=2.17 m$

Rounding off

$d= 2.2 m$

3 0

3 years ago

Dopamine is __________ .

Cerrena [4.2K]

The answer to your question is D.

6 0

3 years ago

Read 2 more answers

Speed<br> =<br> distance = 400m; time = 20s

Natalka [10]

?.............................

6 0

4 years ago

Convert 3598grams into pounds

Sophie [7]

The answer is 7.93 pounds

8 0

3 years ago

Read 2 more answers

Seismographs measure the arrival times of earthquakes with a precision of 0.125 s. To get the distance to the epicenter of the q

zloy xaker [14]

Answer:

435 m

Explanation:

The precision with which the distance to the source of the earthquake can be estimated is equal to the difference in distance covered by the S- and P-waves in the time of 0.125 s.

The distance covered by each type of wave is given by

$d=vt$

where

v is the speed of the waves

t is the time

For S-waves,

v = 3.74 km/s

t = 0.125 s

So the distance covered is

$d_s=(3.74)(0.125)=0.4675 km = 467.5 m$

For P-waves,

v = 7.22 km/s

t = 0.125 s

So the distance covered is

$d_p=(7.22)(0.125)=0.9025 km = 902.5 m$

So, the precision with which the distance can be determined is:

$\Delta d = 902.5 - 467.5 =435 m$

4 0

3 years ago