Ask question

Ask question

All categories

4 years ago

10

An object is thrown directly up (positive direction) with a velocity (vo) of 20.0 m/s and do= 0. How high does it rise (v = 0 cm

/s at top of rise). Remember, acceleration is -9.80 m/s2.
d = _________ m

A. 1.02
B. 2.04
C. 20.4
D. 40.8

1 answer:

Anit [1.1K]4 years ago

4 0

Answer:

The value of d is 20.4 m.

(C) is correct option.

Explanation:

Given that,

Initial velocity = 20 m/s

Final velocity = 0

We need to calculate the time

Using equation of motion

$v = u+gt$

Where, u = Initial velocity

v = Final velocity

Put the value into the formula

$t = \dfrac{20}{9.8}$

$t= 2.04\ sec$

We need to calculate the distance

Using equation of motion

$s = ut+\dfrac{1}{2}gt^2$

$s=0+\dfrac{1}{2}\times9.8\times(2.04)^2$

$s=20.4\ m$

Hence, The value of d is 20.4 m.

You might be interested in

Physical science!!!!!!helpppp

ipn [44]

Answer:

Option B

10.36 m/s

Explanation:

Using the first given equation, then velocity=distance/time

Since distance is provided as 200 m and time, t is 19.3 seconds then substituting these figures yields

v=200/19.3=10.3626943

Rounding off to 2 decimal places, then

v=10.36 m/s

6 0

3 years ago

a ferry which runs at 12m/s in still water travels between towns a and b on a river which flows south at 9 m/s. the ferry leavs

Reil [10]

Answer:

= 15 m/s

Explanation:

Considering right side(west) as positive x-axis and south as negative y-axis.

velocity of boat in still water $v_b=12\hat{i}$

velocity of stream $v_s=-9\hat{j}$

now relative velocity of boat w.r.t. stream $v_{b/s}=12\hat{i}+9\hat{j}$

this velocity with which ac distance will be covered.

therefore magnitude of $v_{b/s} =\sqrt{12^2+9^2}$

= 15 m/s

6 0

3 years ago

Which of the following substances is covalently bonded? A. KF B. H2S C. CaCl2 D. SO2

poizon [28]

The answer is c for that question

6 0

4 years ago

Read 2 more answers

A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that

Marat540 [252]

Answer:

Part(a): The value of the spring constant is $3.11 \times 10^{2}~Kg~s^{-2}$ .

Part(b): The work done by the variable force that stretches the collagen is $1.5 \times 10^{-6}~J$ .

Explanation:

Part(a):

If ' $k$ ' be the force constant and if due the application of a force ' $F$ ' on the collagen ' $\Delta l$ ' be it's increase in length, then from Hook's law

$F = k~\Delta l....................................................................(I)$

Also, Young's modulus of a material is given by

$Y = \dfrac{F/A}{\Delta l/l}...............................................................(II)$

where ' $A$ ' is the area of the material and ' $l$ ' is the length.

Comparing equation ( $I$ ) and ( $II$ ) we can write

$&& Y = \dfrac{l~k}{A}\\&or,& k = \dfrac{Y~A}{l}\\&or,& k = \dfrac{Y~(\pi~r^{2})}{l}$

Here, we have to consider only the circular surface of the collagen as force is applied only perpendicular to this surface.

Substituting the given values in equation ( $III$ ), we have

$k = \dfrac{3.10 \times 10^{6}~N~m^{-2} \times \pi \times (0.00093)^{2}~m^{2}}{.027~m} = 3.11 \times 10^{2}~Kg~s^{-2}$

Part(b):

We know the amount of work done ( $W$ ) on the collagen is stored as a potential energy ( $U$ ) within it. Now, the amount of work done by the variable force that stretches the collagen can be written as

$W = \dfrac{1}{2}k~x^{2} = \dfrac{(\dfrac{F}{k})^{2}k}{2} = \dfrac{F^{2}}{2~k}...................................(IV)$

Substituting all the values, we can write

$W = \dfrac{(3.06 \times 10^{-2})^{2}~N^{2}}{2 \times 3.11 \times 10^{2}~Kg~s^{-2}} = 1.5 \times 10^{-6}~J$

3 0

3 years ago

Is there any other number in the periodic table thats the same as silicon?

barxatty [35]

<span>Astatine same as silicon</span>

8 0

4 years ago

Read 2 more answers