If a melon has a a mass of 1 kg, how much does the melon weigh?

The density of air is 1.3 kg/m^3. What mass of air is contained in a room measuring 2.5m x 4m x 10m? Give your answer in kg. Bra

krek1111 [17]

Mass= density x volume
1.3 kg/m^3 x ( 2.5x4x10) m^3
= 130 kg

7 0

3 years ago

In the context of energy transfers with hot and cold reservoirs, the sign convention is that _______________.

Likurg_2 [28]

Answer:

B. QC > 0; QH < 0

Explanation:

Given that there are two reservoir of energy.

Sign convention for heat and work :

1.If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.

2. If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.

From hot reservoir heat is going out that is why it is taken as negative

$Q_H$

From cold reservoir heat is coming inside the reservoir that is why it is taken as positive

$Q_C>0$

That is why the answer will be

$Q_H$ , $Q_C>0$

8 0

3 years ago

Vector A has a magnitude of 50 units and points in the positive x direction. A second vector, B , has a magnitude of 120 units a

Alex Ar [27]

A) Vector A

The x-component of a vector can be found by using the formula

$v_x = v cos \theta$

where

v is the magnitude of the vector

$\theta$ is the angle between the x-axis and the direction of the vector

- Vector A has a magnitude of 50 units along the positive x-direction, so $\theta_A = 0^{\circ}$ . So its x-component is

$A_x = A cos \theta_A = (50) cos 0^{\circ}=50$

- Vector B has a magnitude of 120 units and the direction is $\theta_B = -70^{\circ}$ (negative since it is below the x-axis), so the x-component is

$B_x = B cos \theta_B = (120) cos (-70^{\circ})=41$

So, vector A has the greater x component.

B) Vector B

Instead, the y-component of a vector can be found by using the formula

$v_y = v sin \theta$

Here we have

- Vector B has a magnitude of 50 units along the positive x-direction, so $\theta_A = 0^{\circ}$ . So its y-component is

$A_y = A sin \theta_A = (50) sin 0^{\circ}=0$

- Vector B has a magnitude of 120 units and the direction is $\theta_B = -70^{\circ}$ , so the y-component is

$B_y = B sin \theta_B = (120) sin (-70^{\circ})=-112.7$

where the negative sign means the direction is along negative y:

So, vector B has the greater y component.

8 0

3 years ago

The wavelength of light that has a frequency of 1.20 × 1013 s-1 is ________ m.

klio [65]

The relationship between frequency and wavelength for an electromagnetic wave is
$c=f \lambda$
where
f is the frequency
$\lambda$ is the wavelength
$c=3 \cdot 10^8 m/s$ is the speed of light.

For the light in our problem, the frequency is $f=1.20 \cdot 10^{13} s^{-1}$ , so its wavelength is (re-arranging the previous formula)
$\lambda= \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{1.20 \cdot 10^{13} s^{-1}}= 2.5 \cdot 10^{-5}m$

8 0

3 years ago

Which sound wave-object interaction is used by animals in echolocation?

tekilochka [14]

The answer is reflection

5 0

3 years ago

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