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3 years ago

What is the acceleration of a dog that runs from 3 m/s to 6 m/s over a distance of 90m?

Physics

1 answer:

KonstantinChe [14]3 years ago

6 0

Answer:

solution given:

acceleration (a)=?

initial velocity (u)=3m/s

final velocity (v)=6m/s

distance (s)=90m

we have

v²=u²+2as

substituting value

6²=3²+2*a*90

36=9+180a

36-9=180a

a=25/180

<u>a=0.1388m/s²</u>

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A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.

AveGali [126]

Answer:

The magnitude of electric force is $7.2\times10^{-3} N$

Explanation:

Coulomb's Law:

The force of attraction or repletion is

directly proportional to the products of charges i.e $F\propto q_1q_2$

inversely proportional to the square of distance i.e $F\propto \frac{1}{r^2}$

$\therefore F\propto \frac{q_1q_2}{r^2}$

$\Rightarrow F=k \frac{q_1q_2}{r^2}$ [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, $r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5$

we know,

$cos \theta = \frac{base }{hypotenuse}$

$\Rightarrow cos \theta = \frac{4 }{r}$

Total force $F_Q = 2.F_1 cos\theta \hat{i}$

$=2k\frac{Qq_1}{r^2} cos\theta \hat i$

$=2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i$

$=8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i$ [ r=5]

$=7.2\times10^{-3}\hat i$ N

The magnitude of electric force is $7.2\times10^{-3} N$

3 0

3 years ago

A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s.

SVEN [57.7K]

Answer:

a) Knowns, initial speed $v_{i}=13.0 m/s$ , final speed $v_{f}=0 m/s$ and gravity due it is a constant $g=9.8m/s^{2}$

b) The maximum high reached by the dolphin is $y_{max}=8.62 m$

c) Total time is $t=2.65s$

Explanation:

a) First of all the initial speed is given at the start of the problem, gravity is constant and final speed is known any object thrown straight up reaches its max high at $0m/s$ speed.

b) Second, now that we know final speed we use $v_{f} =v_{i}-gt$ , as we clear for $t=\frac{v_{i}-v_{f} }{g}=\frac{20.0m/s}{9.8m/s^{2} }=1.32 s$ .

Then we use $y=v_{i}t-\frac{1}{2} gt^{2}=(20.0m/s)(1.32s)-\frac{1}{2} (9.8m/s^{2} ) (1.32s)^{2} =8.62m$

c)Third, finally we can use $y=v_{i}t-\frac{1}{2} gt^{2}$ , as we know $y=0m$ when the dolphin fall into the water again and $v_{i} =13.0m/s$ , then we have $0=(13m/s)t-\frac{1}{2} (9.8m/s^{2} )t^{2}$ is a quadratic form $0=t(13.0-4.9t)$ so we have $t_{i}=0s$ and $t_{f}=\frac{13}{4.90} =2.65s$

6 0

3 years ago

Elias observed a sample in the classroom. The sample was a liquid at room temperature. He performed a conductivity test and foun

Liula [17]

Answer:

Explanation:

Nonmetals are poor conduster to heat and electricity

7 0

3 years ago

Which is true about Pluto

andrew-mc [135]

Answer:

D.)it orbits near the Kepler belt

Explanation:

The Kuiper belt is an area similar to the asteroide belt extending from the orbit of Neptune to about 50 AU from the Sun. It mainly consists of icy asteroids and dwarf planets, which are rocky objects big enough to be defined as planet but that do not have enough gravity to clear their orbit from other obejcts.

Pluto was discovered in 1930 - initially it was classified as a planet, although it is much smaller than the other 8 planets of the Solar System. However, it has been recently de-classified to dwarf planet because its gravity is not enough to clear its orbit from other objects (asteroids). Pluto is located inside the Kuiper belt, so option D is correct. Other dwarf planets in the Kuiper belt are for instance Haumea and Makemake.

8 0

3 years ago

Read 2 more answers

A train slows from 44 m/s to 19 m/s in 9 seconds. What is the train's acceleration? Show your work.

Korvikt [17]

Answer:

a = 2.77 [m/s²]

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f} =v_{o} -a*t$

where:

Vf = final velocity = 19 [m/s]

Vo = initial velocity = 44 [m/s]

a = acceleration [m/s²]

t = time = 9 [s]

Note: The negative sign in the above equation means that the train is decreasing its velocity.

19 = 44 - a*9

9*a = 44 -19

a = 2.77 [m/s²]

3 0

3 years ago