a planes average speed between two cities is 600 km/hr. if he takes 2.5 hrs. how far does the plane fly

Bagaimana cara untuk melatih kemampuan melempar tangkap yang baik dalam bola basket

anyanavicka [17]

Answer:

pergi ke pertandingan sepak bola dan amati

7 0

2 years ago

A system that can be affected by the outside environment, by an exchange of matter or energy is ______.

larisa86 [58]

$\sf{Answer:}$

A system that can be affected by the outside environment, by an exchange of matter or energy is an open physical system .

4 0

2 years ago

A skydiver falls toward the ground at a constant velocity. Which statement best applies Newton’s laws of motion to explain the s

vlabodo [156]

Answer:

The answer is (a.) An upward force balances the downward force gravity on the skydiver

The skydiver is falling at a constant velocity because the upward force is balancing the downward force of gravity. According to Newton, the opposite force balance each other. This is stated in Newton's second law of motion.

8 0

2 years ago

Read 2 more answers

When water freezes, its volume increases by 9.05%. What force per unit area is water capable of exerting on a container when it

AnnZ [28]

Answer:

The Pressure is 0.20 MPa.

(b) is correct option.

Explanation:

Given that,

Change in volume = 9.05%

{tex]\dfrac{\Delta V}{V_{0}}=0.0905[/tex]

We know that.

The bulk modulus for water

$B=0.20\times10^{10}\ N/m^2$

We need to calculate the pressure difference

Using formula bulk modulus formula

$B=\Delta P\dfrac{V_{0}}{\Delta V}$

$\Delta P=B\dfrac{\Delta V}{V_{0}}$

$\Delta P=0.2\times10^{10}\times0.0905$

$\Delta P=0.2\times10^{6}\ Pa$

$\Delta P=0.20 MPa$

Hence, The Pressure is 0.20 MPa.

6 0

3 years ago

A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top o

masha68 [24]

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let <em>T</em> represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, $F_{f2}$ = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = $F_{f2}$ × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

$T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27$

$Frictional \ force, F_{f2} = \left (981 - \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times 0.25 \right) \approx 219.92$