In which situation does heating a substance generally change the temperature?

ladessa [460]

Choices 1, 2, and 4 . . . . . Yes

Choices 3 and 5 . . . . . No

6 0

3 years ago

Researchers at the University of Georgia have evaluated trends in streamside forests in areas within roughly 400 feet of the sta

Colt1911 [192]

B: adds aesthetic value to the landscape. Think about it, out of all your options, that's the one that doesn't really help anything.

And I took the test, so take my word for it.

6 0

2 years ago

A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev

tatuchka [14]

Answer:

$W_f = 2.319 rad/s$

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

$L_i = L_f$

so:

$I_mW_m = I_sW_f$

where $I_m$ is the moment of inertia of the merry-go-round, $W_m$ is the initial angular velocity of the merry-go-round, $I_s$ is the moment of inertia of the merry-go-round and the child together and $W_f$ is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = $\frac{1}{2}M_mR^2$

I = $\frac{1}{2}(115 kg)(2.5m)^2$

I = 359.375 kg*m^2

Where $M_m$ is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2 $\pi$ rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

$I_s = \frac{1}{2}M_mR^2+mR^2$

$I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2$

$I_s = 506.25kg*m^2$

Finally we replace all the data:

$(359.375)(3.2672) = (506.25)W_f$

Solving for $W_f$ :

$W_f = 2.319 rad/s$

7 0

2 years ago

A bowling ball of mass 5 kg rolls down a slick ramp 20 meters long at a 30 degree angle to the horizontal. What is the work done

Elena-2011 [213]

Answer:

The work done by gravity during the roll is 490.6 J

Explanation:

The work (W) is:

$W = F*d$

Where:

F: is the force

d: is the displacement = 20 m

The force is equal to the weight (W) in the x component:

$F = W_{x} = mgsin(\theta)$

Where:

m: is the mass of the bowling ball = 5 kg

g: is the gravity = 9.81 m/s²

θ: is the degree angle to the horizontal = 30°

$F = mgsin(\theta) = 5 kg*9.81 m/s^{2}*sin(30) = 24.53 N$

Now, we can find the work:

$W = F*d = 24.53 N*20 m = 490.6 J$

Therefore, the work done by gravity during the roll is 490.6 J.

I hope it helps you!

6 0

2 years ago

What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of – 30.0 nC?

Nastasia [14]

Answer:

$F=1.26*10^{-3}N$

Explanation:

Assuming the pith balls as point charges, we can calculate the repulsive force between them, using Coulomb's law:

$F=\frac{kq_1q_2}{d^2}$

We observe that the magnitude of the electric force is directly proportional to the product of the magnitude of both signed charges( $q_1,q_2$ ) and inversely proportional to the square of the distance(d) that separates them.

Replacing the given values, where k is the Coulomb constant:

$F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-30*10^{-9}C)(-30*10^{-9}C)}{(8*10^{-2}m)^2}\\F=1.26*10^{-3}N$

8 0

2 years ago