What principal of a lever is being used in this photograph?

Your GPS shows that your friend’s house is 10.0 km away. But there is a big hill between your houses and you don’t want to bike

kati45 [8]

Answer:

The value is $c = 8 \ km$

Explanation:

From the question we are told that

The distance of friends house from your point is $a = 10 \ km$

The distance of your friends street from your street is $b = 6 \ km \ in the \ direction \ towards \ the \ north$

The diagram illustrating this question is shown on the first uploaded image

From the diagram we can apply by Pythagoras theorem as follows

$a^2 = b^2 + c^2$

=> $c = \sqrt{^2 - b^2}$

=> $c = \sqrt{ 10^2 - 6^2}$

=> $c = 8 \ km$

6 0

3 years ago

How would you best define moving charges

svet-max [94.6K]

Current.A moving charge is an object that changes position to one particular obsever.

8 0

3 years ago

Sarah weighs 500 newtons. She climbs a set of stairs 6 meters high. How much work does she do? 83 J Work equals force times dist

jek_recluse [69]

Answer:

<em>Answer: Work equals force times distance. 3,000 J</em>

Explanation:

Work Done By A Force

When some force $\vec F$ is applied and a displacement $\vec r$ is achieved, the work done by the force is given by

$W=\vec F \cdot \vec r$

Note that the work is a scalar magnitude as the result of the dot-product of two vectors. If the force and the displacement are parallel, then the vectors can be replaced as its magnitudes F,x and the work is

$W=F.x$

The dot product becomes a simple arithmetic product, i.e force times distance.

Sara weighs 500 Nw and she climbs up a 6 meter set of stairs. She needs to lift her weight up, so the force is the weight and the distance is the height of the stairs, thus

$W=500\ Nw.\ 6\ m=3,000\ Joule$

Answer: Work equals force times distance. 3,000 J

7 0

3 years ago

Students are given the following information about a 2.0 kg, motorized toy boat on water. what net force is exerted on the boat

Natasha_Volkova [10]

The motorized toy boat experiences a net force of 0 N between 4 s and 8 s.

The motorized toy boat moves at 8 m/s (u) at 4 s and at 8 m/s (v) at 8 s. We can calculate the acceleration (a) in that period using the following kinematic expression.

$a = \frac{v-u}{\Delta t} = \frac{8m/s-8m/s}{8s-4s} = 0 m/s^{2}$

The object with a mass (m) of 2.0 kg experiences an acceleration of 0 m/s². We can calculate the net force (F) in that period using Newton's second law of motion.

$F = m \times a = 2.0 kg \times 0 m/s^{2} = 0 N$