The map here shows land surface temperatures on earth during one month in 2015

Stels [109]

Answer:

An electric bell is placed inside a transparent glass jar. The bell can be turned on and off using a switch on the outside of the jar. A vacuum is created inside the jar by sucking out the air. Then the bell is rung using the switch. What will we see and hear?

A.

We’ll see the bell move, but we won’t hear it ring.

B.

We won’t see the bell move, but we’ll hear it ring.

C.

We’ll see the bell move and hear it ring.

D.

We won’t see the bell move or hear it ring.

E.

We’ll see the sound waves exit the vacuum pump.

Explanation:

so, the answer to the question is

A.

We'll see the bell move, but we won’t hear it ring.

5 0

2 years ago

Read 2 more answers

Consider a projectile of mass 20 kg launched with a speed 9 m/s at an elevation angle of 45 degrees. Taking the launch point as

viva [34]

Answer:

a) L=0. b) L = 262 k ^ Kg m²/s and c) L = 1020.7 k^ kg m²/s

Explanation:

It is angular momentum given by

L = r x p

Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum

One of the easiest ways to make this vector product is with the use of determinants

${array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]$

Let's apply this relationship to our case

Let's start by breaking down the speed

v₀ₓ = v₀ cosn 45

voy =v₀ sin 45

v₀ₓ = 9 cos 45

voy = 9 without 45

v₀ₓ = 6.36 m / s

voy = 6.36 m / s

a) at launch point r = 0 whereby L = 0

. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero

vfy² = voy²- 2 g y

y = voy² / 2g

y = (6.36)²/2 9.8

y = 2.06 m

Let's calculate the angular momentum

L= $\left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]$

L = -px y k ^

L = - (m vox) (2.06) k ^

L = - 20 6.36 2.06 k ^

L = 262 k ^ Kg m² / s

The angular momentum is on the z axis

c) At the point of impact, at this point the height is zero and the position on the x-axis is the range

R = vo² sin 2θ / g

R = 9² sin (2 45) /9.8

R = 8.26 m

L = $\left[\begin{array}{ccc}i&j&k\\x&0&0\\px&py&0\end{array}\right]$

L = - x py k ^

L = - x m voy

L = - 8.26 20 6.36 k ^

L = 1020.7 k^ kg m² /s

5 0

3 years ago

A blank is a compound that turns litmus blue and is often found in soaps and detergents

dangina [55]

An ... (Base ) .... is a compound that turns red litmus paper blue and is often found in soaps and detergents.

7 0

2 years ago

A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.

kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

$W=Fd cos \theta$

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

$\theta=30^{\circ}$ is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

$W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J$

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

$\Delta K = W$

where

$\Delta K$ is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

$\Delta K=69.3 J$

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

$\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, $\Delta K$ , after it has travelled for d=3 m. Using the work-energy theorem again, we find

$\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J$

And substituting into (1) and re-arrangin the equation, we find

$v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s$

6 0

3 years ago

In the pair of supply and demand equations below, where x represents the quantity demanded in units of a thousand and p the unit

arlik [135]

Answer:

Equilibrium quantity = 5

Equilibrium price = 40

Explanation:

given:

p = -x²-3x+80

p = 7x+5

For the equilibrium quantity the price from both the functions will be equal

thus, we have

-x² - 3x + 80 = 7x+5

⇒ x² +3x + 7x + 5 - 80 = 0

⇒x² + 10x - 75 = 0

now solving for x

x²- 5x + 15x -75 = 0

x(x-5) + 15(x-5) = 0

therefore, the two roots of the equation are

x = 5 and x = -15

since the quantity cannot be in negative

therefore, the equilibrium quantity will be = 5

now the equilibrium price can be found out by substituting the equilibrium quantity in any of the equation

thus,

p = -(5)² -3(5) + 80 = 40

or

p = 7(5) + 5 = 40

4 0

2 years ago