All categories

1 year ago

6. A student travels 4.0 m east, 8.0 m south, 4.0 m west, and finally 8.0 m north. What is her

Physics

1 answer:

Helen [10]1 year ago

3 0

The student's path traces out a rectangle and she ends up at the same spot, so her <u>displacement</u> is 0 m.

On the other hand, she travels a <u>total distance</u> equal to the perimeter of that rectangle,

4.0 m + 8.0 m + 4.0 m + 8.0 m = 24 m

You might be interested in

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

2 years ago

Milk containing 3.7% fat and 12.8% total solids is to be evaporated to produce a product containing 7.9% fat. What is the yield

ratelena [41]

Answer:

the yield of product is YP=46.835 % and the concentration of solids is

Cs = 27.33%

Explanation:

Assuming that all the solids and fats remains in the milk after the evaporation, then the mass of product mP will be

Mass of fat in 100 kg of milk = 100 kg* 0.037 = mP* 0.079

mP = 100 kg* 0.037/0.079 = 46.835 kg

then the yield YP of the product is

YP= mP / 100 kg = 46.835 kg / 100 kg = 46.835 %

YP= 46.835 %

the concentration of solids Cs is

Mass of solids in 100 kg of milk = 100 kg* 0.128 = 46.835 kg * Cs

Cs = 100 kg* 0.128 / 46.835 kg = 0.2733 = 27.33%

Cs = 27.33%

3 0

2 years ago

Suppose that two point charges, each with a charge of +1.00 C, are separated by a distance of 1.0 m. If the distance between the

Aleksandr-060686 [28]

Given:

The magnitude of each charge is q1 = q2 = 1 C

The distance between them is r = 1 m

To find the force when distance is doubled.

Explanation:

The new distance is

$\begin{gathered} r^{\prime}=\text{ 2r} \\ =2\times1 \\ =2\text{ }m \end{gathered}$

The force can be calculated by the formula

$F=k\frac{q1q2}{(r^{\prime})^2}$

Here, k is the constant whose value is

$k=9\times10^9\text{ N m}^2\text{ /C}^2$

On substituting the values, the force will be

$\begin{gathered} F=9\times10^9\times\frac{1\times1}{(2)^2} \\ =2.25\times10^9\text{ N} \end{gathered}$

7 0

1 year ago

Two ships of equal mass are 110 m apart. What is the acceleration of either ship due to the gravitational attraction of the othe

dusya [7]

Answer:

Acceleration of the ship, $a=2.14\times 10^{-7}\ m/s^2$

Explanation:

It is given that,

Mass of both ships, $m=39000\ metric\ tons=39\times 10^6\ kg$

Distance between two ships, d = 110 m

The gravitational force between two ships is given by :

$F=G\dfrac{m^2}{d^2}$

$F=6.67\times 10^{-11}\ Nm^2/kg^2\times \dfrac{(39\times 10^6\ kg)^2}{(110\ m)^2}$

F = 8.38 N

Let a is the acceleration. Now, using second law of motion as :

$a=\dfrac{F}{m}$

$a=\dfrac{8.38\ N}{39\times 10^6\ kg}$

$a=2.14\times 10^{-7}\ m/s^2$

So, the acceleration of either ship due to the gravitational attraction of the other is $2.14\times 10^{-7}\ m/s^2$ . Hence, this is the required solution.

7 0

2 years ago

What percent of sample of AS-198 to decay to 1/8 its original

denis-greek [22]

Answer:

bannana

Explanation:

5 0

3 years ago