There is a bell at the top of a tower that is 45 meters high. The bell weighs 210 N. The

A house has well-insulated walls. It contains a volume of 105 m3 of air at 305 K.

REY [17]

Answer: $85.46\ kJ$

Explanation:

Given

Volume of air $V=105\ m^3$

Temperature of air $T=305\ K$

Increase in temperature $\Delta T=0.7^{\circ}C$

Specific heat for diatomic gas is $C_p=\dfrac{7R}{2}$

Energy required to increase the temperature is

$\Rightarrow Q=nC_pdT\\\\\Rightarrow Q=n\times \dfrac{7R}{2}\times \Delta T\\\\\Rightarrow Q=\dfrac{7}{2}nR\Delta T\\\\\Rightarrow Q=\dfrac{7}{2}\times \dfrac{PV}{T}\times \Delta T\quad [\text{using PV=nRT}]$

Insert the values

$\Rightarrow Q=\dfrac{7}{2}\times \dfrac{1.01325\times 10^5\times 105}{305}\times 0.7\\ \text{Assuming air pressure to be atmospheric P=}1.01325\times 10^5\ N/m^2\\\\\Rightarrow Q=0.8546\times 10^5\\\Rightarrow Q=85.46\ kJ$

6 0

3 years ago

What happens to a gas when its temperature decreases?

TiliK225 [7]

Hey There!

When the gas temperature decreases the volume decreases too.

Thank You!

Have a question? Need Support? Feel free to message me.

6 0

3 years ago

Read 2 more answers

The difference between the observed points and the regression line points is equal to the:________

pshichka [43]

The difference between the observed points and the regression line points is equal to the correlation.

The strength and direction of a relationship between two or more variables are described by the statistical measure of correlation, which is given as a number. However, a correlation between two variables does not necessarily imply that a change in one variable is the reason for a change in the values of the other.

Regression expresses the relationship as an equation, whereas correlation assesses the strength of the linear link between two variables. The square of the correlation coefficient, also known as Pearson's r, between the observed and predicted values in a regression is sometimes referred to as R2.

Learn more about correlation here;

brainly.com/question/6563788

#SPJ4

8 0

2 years ago

Behaviors in deer such as herding and male deer using antlers to compete for females increase the chances of successful reproduc

lorasvet [3.4K]

Answer:

True

Explanation:

The reproductive success of any species is the capability to produce their offspring per breeding lifetime or event.

Most of the species have to attract their partners by their physical capability and build up so that the mother can choose her partner in order to breed the best kind of off spring.

In the context, in case of deer, the size of the their antlers as well as behavior in herding is considered as the best chances for a successful reproduction to compete among the males and find their breeding mate.

Thus the answer is TRUE.

6 0

2 years ago

A 2028 kg Oldsmobile traveling south on Abbott Road at 14.5 m/s is unable to stop on the ice covered intersection for a red ligh

Svetradugi [14.3K]

Answer:

v = 8.1 m/s

θ = -36.4º (36.4º South of East).

Explanation:

Assuming no external forces acting during the collision (due to the infinitesimal collision time) total momentum must be conserved.
Since momentum is a vector, if we project it along two axes perpendicular each other, like the N-S axis (y-axis, positive aiming to the north) and W-E axis (x-axis, positive aiming to the east), momentum must be conserved for these components also.
Since the collision is inelastic, we can write these two equations for the momentum conservation, for the x- and the y-axes:
We can go with the x-axis first:

$p_{ox} = p_{fx} (1)$

⇒ $m_{tr} * v_{tr}= (m_{olds} + m_{tr}) * v_{fx} (2)$

Replacing by the givens, we can find vfx as follows:

$v_{fx} = \frac{m_{tr}*v_{tr} }{(m_{tr} + m_{olds)} } = \frac{4146kg*9.7m/s}{2028kg+4146 kg} = 6.5 m/s (3)$

We can repeat the process for the y-axis:

$p_{oy} = p_{fy} (4)$

⇒ $m_{olds} * v_{olds}= (m_{olds} + m_{tr}) * v_{fy} (5)$

Replacing by the givens, we can find vfy as follows:

$v_{fy} = \frac{m_{olds}*v_{olds} }{(m_{tr} + m_{olds)} } = \frac{2028kg*(-14.5)m/s}{2028kg+4146 kg} = -4.8 m/s (6)$

The magnitude of the velocity vector of the wreckage immediately after the impact, can be found applying the Pythagorean Theorem to vfx and vfy, as follows:

$v_{f} = \sqrt{v_{fx} ^{2} +v_{fy} ^{2} }} = \sqrt{(6.5m/s)^{2} +(-4.8m/s)^{2}} = 8.1 m/s (7)$

In order to get the compass heading, we can apply the definition of tangent, as follows:

$\frac{v_{fy} }{v_{fx} } = tg \theta (8)$

⇒ tg θ = vfy/vfx = (-4.8m/s) / (6.5m/s) = -0.738 (9)

⇒ θ = tg⁻¹ (-0.738) = -36.4º

Since it's negative, it's counted clockwise from the positive x-axis, so this means that it's 36.4º South of East.

5 0

2 years ago