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2 years ago

1) A parked car starts to move with a constant velocity of 3 m/s, after moving

Physics

1 answer:

makkiz [27]2 years ago

7 0

Answer:

velocity (v) =distance(s) ÷ time (t)

Explanation:

v=s÷t

s=v×t

s=3×3.4

s= 10.2m

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After a plant or animal dies the 14C content decreases with a half-life of 5730 years. If an archaeologist finds an ancient fire

Shalnov [3]

Answer:

age of the site is 15411.75 years old

Explanation:

Given data

plant or animal dies = 14C

time period = 5730 year

carbon = 15.5%

to find out

age (in years) of the ancient site

solution

we know that Final value = Initial value × $0.5^{n}$

here n is half life passed

so for 15.5%

15.5% = 100% of $0.5^{n}$

0.155 = 1 × $0.5^{n}$

now take log both side

log 0.155 = log $0.5^{n}$

n = log 0.155 / log 0.5

n = 2.68966

we know here 5730 years in half life

so for 2.68966 half-lives = 2.68966 × 5730 = 15411.7518

age of the site is 15411.75 years old

6 0

2 years ago

A 125 g pendulum bob hung on a string of length 35 cm has the same period as when the bob is hung from a spring and caused to os

Bingel [31]

Answer:

k = 3.5 N/m

Explanation:

It is given that the time period the bob in pendulum is the same as its time period in spring mass system:

$Time\ Period\ of\ Pendulum = Time\ Period\ of\ Spring-Mass\ System\\2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{m}{k}$

$\frac{l}{g} = \frac{m}{k}\\\\ k = g\frac{m}{l}$

where,

k = spring constant = ?

g = acceleration due to gravity = 9.81 m/s²

m = mass of bob = 125 g = 0.125 kg

l = length of pendulum = 35 cm = 0.35 m

Therefore,

$k = (9.81\ m/s^2)(\frac{0.125\ kg}{0.35\ m})\\\\$

4 0

2 years ago

Are cells made of tisse?

zepelin [54]

Answer:

Yes

Explanation:

Cells make up tissues. Hope this helped

5 0

2 years ago

Read 2 more answers

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$