All categories

4 years ago

Sound waves require a medium to travel through, such as a solid, liquid or gas because

Physics

2 answers:

Natasha2012 [34]4 years ago

8 0

A

because waves need air or something else to travel through

Katyanochek1 [597]4 years ago

8 0

D sound is molecules bumping into one another

You might be interested in

Many things can alter your heart rate including: exercise, diet, nutrition, sugar, and caffeine

Dmitry_Shevchenko [17]

Answer:

true

Explanation:

4 0

3 years ago

A good soccer player can kick the ball up to 25 m/sec. A soccer ball has a mass of 800 grams (0.8 kg). What force must a goalie

Oksana_A [137]

Answer:

200 N

Explanation:

Applying,

The force a golie must exert on the ball is,

F = ma...................... Equation 1

Where m = mass of the ball, a = acceleration of the ball.

But,

a = Δv/t............... Equation 2

Where Δv = change in velocity, t = time.

Substitute equation 2 into equation 1

F = m(Δv/t)............... Equation 3

From the question,

Given: m = 0.8 g, t = 0.1 s, Δv = 25 m/s

Substitute these values into equation 3

F = 0.8×25/0.1

F = 200 N

6 0

3 years ago

The electric field on the surface of an irregularly shaped conductor varies from 74.0 kN/C to 14.0 kN/C.

mrs_skeptik [129]

Answer:

(a). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

Explanation:

Given that,

Electric field $E_{1}=74.0\ kN/C$

Electric field $E_{2}=14.0\ kN/C$

When the radius of curvature is greatest, the electric field at the surface will be smaller.

Where the radius of curvature is greatest

(a). We need to calculate the local surface charge density at the point on the surface

Using formula of charge density

$\sigma=\epsilon_{0}E_{2}$

Put the value into the formula

$\sigma=8.85\times10^{-12}\times14\times10^{3}$

$\sigma=1.239\times10^{-7}\ C/m^2$

$\sigma=123.9\times10^{-9}\ C/m^2$

$\sigma=123.9\ nC/m^2$

The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). We need to calculate the local surface charge density at the point on the surface where the radius of curvature of the surface is smallest

Using formula of charge density

$\sigma=\epsilon_{0}E_{1}$

Put the value into the formula

$\sigma=8.85\times10^{-12}\times74\times10^{3}$

$\sigma=6.549\times10^{-7}\ C/m^2$

$\sigma=654.9\times10^{-9}\ C/m^2$

$\sigma=654.9\ nC/m^2$

The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

Hence, (a). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

7 0

3 years ago

List at least 5 examples of civil rights

mariarad [96]

Answer:

right to vote, the right to a fair trial, the right to government services, the right to a public education, and the right to use public facilities.

6 0

3 years ago

A 1.0-kg standard cart collides on a low-friction track with cart A. The standard cart has an initial x component of velocity of

Andru [333]

Answer:

Explanation:

Mathematically, linear momentum is expressed as the product of mass and velocity. Linear momentum conservation law states that a body or system of bodies retains its total momentum unless an external force is applied to the system.

In this case, the system consists of two carts.

At the start, the linear momentum (P) of the system is equal to:

$P=1.0kg*0.4m/s=0.4kg*m/s$

It's only composed of linear momentum of the standard cart because cart A doesn't have any linear momentum at that moment.

After the collision, linear momentum has to be the same

$P=0.4kg*m/s=1.0kg*0.20m/s+m_{A} *0.70m/s$

where m_A is the mass of the cart A.

Solving for m_A

$m_{A} =0.28kg$

After the cart A rebounds, the linea momentum of the system has changed (because of the force present in the rebound). The new linear momentum is:

$P=1kg*0.2m/s+0.29kg*(-0.7m/s)=-0.003kg*m/s$

Then, the lump of putty is added to the system, but the linear momentum has to be the same, because we added a mass, not a force. The mass of that putty (m_p) has to be added to the equation of the system

$-0.003kg*m/s=1.0kg*(-0.2m/s)+(0.29kg+m_{p} )(0.4m/s)$

Solving for m_p

$m_{p}=0.20kg$

6 0

3 years ago