1) An object travels 15 m in 3 s. What is its' speed?

A large piece of jewelry has a mass of 132.6 g. a graduated cylinder initially contains 48.6 ml water. when the jewelry is subme

Lerok [7]

The density of the substance is 10.5 g/cm³.

The jewelry is made out of Silver.

Density ρ is defined as the ratio of mass m of the substance to its volume V. The cylinder contains a volume V₁ of water and when the jewelry is immersed in it, the total volume of water and the jewelry is found to be V₂.

The volume V of the jewelry is given by,

$V=V_2 -V_1$

Substitute 48.6 ml for V₁ and 61.2 ml for V₂.

$V=V_2 -V_1\\ =61.2 ml -48.6 ml\\ =12.6 ml$

calculate the density ρ of the jewelry using the expression,

$\rho =\frac{m}{V}$

Substitute 132.6 g for m and 12.6 ml for V.

$\rho =\frac{m}{V}\\ =\frac{132.6 g}{12.6 ml} \\ =10.5 g/ml$

Since $1 ml=1 cm^3$ ,

The density of the jewelry is 10.5 g/cm³.

From standard tables, it can be seen that the substance used to make the jewelry is silver, which has a density 10.5 g/cm³.

4 0

3 years ago

What is the unit for resistance?

AnnyKZ [126]

For electrical resistance, the unit is Ohm. The symbol is capital greek O.

4 0

3 years ago

A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the a

stira [4]

Answer:

Explanation:

Given

acceleration is given by

$a=-g-c_Dv^2$

where $\ddot{y}=a$

$\dot{y}=v$

Also acceleration is given by

$a=v\frac{\mathrm{d} v}{\mathrm{d} s}$

$ds=\frac{v}{a}dv$

$\int ds=\int \frac{v}{-g-0.001v^2}dv$

$\Rightarrow Let -g-0.001v^2=t$

$-0.001\times 2vdv=dt$

$vdv=-\frac{dt}{0.002}$

$at\ v_0=50\ m/s,\ t=-g-0.001(50)^2$

$t=-g-2.5$

at $v=0,\ t=-g$

$\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}$

$\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}$

$s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}$

$s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})$

$s=113.608\ m$

when air drag is neglected maximum height reached is

$h=\frac{v_0^2}{2g}$

$h=\frac{50^2}{2\times 9.8}$

$h=127.55\ m$

3 0

3 years ago

Last night Mookie Betts hit a baseball at 32.5 m/s at a 45° angle. Betts

podryga [215]

Answer:

a) Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) The total distance traveled by the baseball was 108.7 m.

Explanation:

a) To know if the hit was a home run we need to calculate the height of the ball at 99 m:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final height =?

$y_{0}$ : is the initial height = 1 m

$v_{0_{y}$ : is the initial vertical velocity = v₀sin(45)

v₀: is the initial velocity = 32.5 m/s

g: is the gravity = 9.81 m/s²

t: is the time

First, we need to find the time by using the following equation:

$t = \frac{x}{v_{0_{x}}} = \frac{99 m}{32.5 m/s*cos(45)} = 4.31 s$

Now, the height is:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2} = 1m + 32.5 m/s*sin(45)*4.31 s - \frac{1}{2}9.81 m/s^{2}*(4.31 s)^{2} = 8.93 m$

Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) To find the distance traveled by the baseball first we need to find the time of flight:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

$0 = 1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2}$

$1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2} = 0$

By solving the above quadratic equation we have:

t = 4.73 s

Finally, with that time we can find the distance traveled by the baseball:

$x = v_{0_{x}}*t = 32.5 m/s*cos(45)*4.73 s = 108.7 m$

Hence, the total distance traveled by the baseball was 108.7 m.

I hope it helps you!

4 0

3 years ago

Which of the following most likely happens when the temperature of a chemical reaction is increased?

Diano4ka-milaya [45]

Molecules Speed Up because the other three options only occur when the temperature is decreased.

5 0

3 years ago

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1) An object travels 15 m in 3 s. What is its' speed?​

1) An object travels 15 m in 3 s. What is its' speed?