Someone please help??

How can the rate of a reaction be increased?

OlgaM077 [116]

The Answer Is Lowering the amount of reactants

7 0

3 years ago

A pilot drops a bomb from a plane flying horizontally with constant velocity. When the bomb hits the ground, the horizontal loca

polet [3.4K]

Answer: The horizontal location of the plane will BE OVER THE BOMB

Explanation:

As soon as the bomb was dropped, the bomb will fall under gravity (free fall) and the location of the plane continues to increase horizontally till the bomb reaches the ground which is a falling distance to be travelled by the bomb at 9.8m/s²

8 0

3 years ago

A 43.9-g piece of copper (CCu= 0.385 J/g°C) at 135.0°C is plunged into 254 g of water at 39.0°C. Assuming that no heat is lost t

Semmy [17]

Answer:

$T = 40.501\,^{\textdegree}C$

Explanation:

The interaction of the piece of copper and water means that the first one need to transfer heat in order to reach a thermal equilibrium with water. Then:

$-Q_{out,Cu} = Q_{in,H_{2}O}$

After a quick substitution, the expanded expression is:

$-(43.9\,g)\cdot (0.385\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-135^{\textdegree}C) = (254\,g)\cdot (4.187\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-39\,^{\textdegree}C)$

$-16.902\,\frac{J}{^{\textdegree}C}\cdot (T-135^{\textdegree}C) = 1063.498\,\frac{J}{^{\textdegree}C} \cdot (T-39^{\textdegree}C)$

$43758,192\,J = 1080.4\,\frac{J}{^{\textdegree}C}\cdot T$

The final temperature of the system is:

$T = 40.501\,^{\textdegree}C$

8 0

3 years ago

Read 2 more answers

A baseball of mass m = 0.31 kg is spun vertically on a massless string of length L = 0.51m. The string can only support a tensio

natulia [17]

Given data:

* The mass of the baseball is 0.31 kg.

* The length of the string is 0.51 m.

* The maximum tension in the string is 7.5 N.

Solution:

The centripetal force acting on the ball at the top of the loop is,

$\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}$

For the maximum velocity of the ball at the top of the vertical circular motion,

$v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}$

where g is the acceleration due to gravity,

Substituting the known values,

$\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}$

Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.

8 0

1 year ago

Please choose the answer that describes the scientific notation for 5,098,000.

galben [10]

D. 5.098 x 106 is the correct answer when reduced to the proper notation.

3 0

3 years ago