Please Help! <br><br> #1 and #2 are the ones i need help on

Claire is traveling home when her mother calls and asks her to go to her grandma's house. She is 300 meters north of her house w

m_a_m_a [10]

Answer:

you divide distance traveled by the time it takes to travel that distance. average speed. If your speed changes from 10 km/h to 6 km/h, you have a(n) ... 400 km. Suppose that the average speed your dog can run is 3 m/s. ... she drives her scooter 7 kilometres north. She stops for lunch and then drives 5 kilometres south.

i rlly dunno lma.o

Explanation:

5 0

3 years ago

Help awnser need experts

pishuonlain [190]

12. The answer would be C. 1.50 s. This is because if you divide 60 by 40, you will get 1.5.

13. For this one I'm not sure, but what I can tell you is that the heavier something is the faster it will sink, the lighter it is, it will float.

5 0

4 years ago

A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 514 N. As the drawing sh

n200080 [17]

Answer:

Part a)

$T_L = 155.4 N$

Part b)

$T_R = 379 N$

Explanation:

As we know that mountain climber is at rest so net force on it must be zero

So we will have force balance in X direction

$T_L cos65 = T_R cos80$

$T_L = 0.41 T_R$

now we will have force balance in Y direction

$mg = T_L sin65 + T_Rsin80$

$514 = 0.906T_L + 0.985T_R$

Part a)

so from above equations we have

$514 = 0.906T_L + 0.985(\frac{T_L}{0.41})$

$514 = 3.3 T_L$

$T_L = 155.4 N$

Part b)

Now for tension in right string we will have

$T_R = \frac{T_L}{0.41}$

$T_R = 379 N$

3 0

3 years ago

A reservoir located in the mountain 250 m above sea level flows through a pipe to a hydroelectric plant in a town at sea level.

Pavlova-9 [17]

Answer:

v₂ = 70 m / s

Explanation:

For this exercise let's use Bernoulli's equation

where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂

indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute

ρ g y₁ = ½ ρ v₂²

v₂ = $\sqrt {2g \ y_1}$

let's calculate

v₂ = √( 2 9.8 250)

v₂ = 70 m / s

6 0

3 years ago

in the figure shown, if the mass of the block is 4kg and the coefficient of static friction is 0.5 and the coefficient of kineti

Elan Coil [88]

Answer:

Ff = 19.6 N

Explanation:

So since its saying whats the minimum F to move the block, we will use static friction (0.5).

We will use the equation for force of friction, which is Ff = uFn

Ff = (0.5)(4)(9.8)

Ff = 19.6 N

this is the minumum force needed to move the block, as that is the frictional force. You would need to apply a minimum force of 19.6 N to move the block

3 0

3 years ago

Please Help! #1 and #2 are the ones i need help on