Is calcium atom an acid or base?

A bus accelerates to 60 m/s to the east in 10 s. What is the buses acceleration? 2. A car traveling at 10.0 m/s to the west acce

professor190 [17]

Answers:

1) $a=6\frac{m}{s^{2}}$

2) $t=8s$

Explanation:

1) Acceleration $a$ is defined as the variation of Velocity $V$ in time $t$ :

$a=\frac{V}{t}$ (1)

A body also has acceleration when it changes its direction.

In this case we have a bus with a velocity of 60m/s to the east, that accelerates in a time 10s. So, we have to find the bus's acceleration:

$a=\frac{60m/s}{10s}$ (2)

$a=6m/s^{2}$ (3) This is the bus's accelerration

2) Now we have a car that accelerates $2m/s^{2}$ to the west in order to reach a speed of $16m/s$ in the same direction, and we have to find the time $t$ it takes to the car to reach that velocity.

Therefore we have to find $t$ from (1):

$t=\frac{V}{a}$ (4)

$t=\frac{16m/s}{2m/s^{2}}$ (5)

Finally:

$t=8s$ (6)

3 0

3 years ago

На рычаге размещены два противовеса таким образом, что рычаг находится в состоянии равновесия. Вес расположенного слева противов

prisoha [69]

Answer:

Вес противовеса, или сила нагрузки, составляет затем 100 000 фунтов-футов, разделенных на 20 футов, или 5000 фунтов.

Explanation:

8 0

3 years ago

A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f

VashaNatasha [74]

Answer:

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 = X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + $(\frac{L-y}{2})$ ) g sinФ - F2 x ( $\frac{L}{2}$ ) x gsinФ = 0

plug in the equations of F1 and F2 into the above equation and after simplification, we get:

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . 5/9 x w

Hence,

$(L^{2} - y^{2} )$ = $\frac{5L^{2} }{9}$

$\frac{L^{2} - y^{2} }{L^{2} }$ = $\frac{5}{9}$

Taking $L^{2}$ common and solving for $\frac{y}{L}$ , we will get

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

8 0

3 years ago

A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica

Ugo [173]

Answer:

$f_{e}$ = 1.9 cm

Explanation:

The magnification of a microscope is the product of the magnification of the eyepiece by the magnifier with the objective

M = M₀ $m_{e}$

Where M₀ is the magnification of the objective and $m_{e}$ is the magnification of the eyepiece.

The eyepiece is focused to the near vision point (d = 25 cm)

$m_{e}$ = 25 / $f_{e}$

The objective is focused on the distances of the tube (L)

M₀ = -L / f₀

Substituting

M = - L/f₀ 25/ $f_{e}$

1) Let's look for the focal length of the eyepiece (faith)

$f_{e}$ = - L 25 / f₀ M

M = 400X = -400

$f_{e}$ = - 12 25 /0.40 (-400)

$f_{e}$ = 1.875 cm

Let's approximate two significant figures

$f_{e}$ = 1.9 cm

8 0

3 years ago

What do you understand by family health?<br>

AleksandrR [38]

Answer:

Family health provides, besides the cure of disease, various activities and programs in prevention, rehabilitation, and health promotion for all members of families/households. pls mark me branilest

3 0

3 years ago

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