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3 years ago

A cat runs at 2 m/s for 3 s and then slows to a stop with an acceleration of 0.80 m/s2. What is

Physics

1 answer:

sp2606 [1]3 years ago

7 0

Answer:

see that the correct one is B

Explanation:

To solve this exercise let us use the kinematic relations

v² = v₀² - 2 a x

as they indicate that the car stops, therefore the final speed is yield v = 0

x = v₀² / 2a

let's calculate

x = 2²/(2 0.8)

x = 2.5 m / s²

When reviewing the answers we see that the correct one is B

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what is the acceleration of a softball if it has a mass of 0.50kg and hits the catcher’s glove with a force of 25 N

Kitty [74]

Answer:

mass=0.50kg

force=25N

acceleration =?

Now,

force=m×a

25=0.50×a

25÷0.50=a

50=a

acceleration =50m/s^2 answer!!!!

hope this may help you!!!!

3 0

3 years ago

In 1996, astronomers discovered an icy object beyond pluto that was given the designation 1996 tl 66. it has a semimajor axis of

antoniya [11.8K]

Answer : 2446 years.

Explanation :

Length of semi major axis is, $a=84\ au= 1.496\times 10^{11}\ m$

According to Kepler's third law, square of time period of an orbit is directly proportional to the cube of the semi major axis.

i.e $T^2=\dfrac{4\pi^2}{GM}a^3$

where G is gravitational constant

M is mass of sun, $M=1.98\times 10^{30}\ Kg$

So, $T^2=\dfrac{4\times (3.14)^2}{6.6\times 10^{-11}Nm^2/Kg\times 1.98\times 10^{30}\Kg}$

$T^2=3\times 10^{-19}\times(84\times 1.496\times 10^{11})^3$

$T^2=3\times 10^{-19}\times 1984415.6\times 10^{33}$

$T^2=59532469.8\times 10^{14}\ s$

$T=7715.7\times 10^7\ s$

since, $1\ sec=3.17\times 10^{-8}\ years$

So, orbital period is approximately 2446 years.

7 0

4 years ago

In an experiment to determine the s.h.c. of lead, a 0.80 kg block of lead is heated using a

gladu [14]

Answer:2.47

Explanation: did the math

8 0

2 years ago

There is a naturally occurring vertical electric field near the Earth’s surface that points toward the ground. In fair weather c

trasher [3.6K]

Answer:

$\frac{F}{W} = 9.37 \times 10^{-4}$

Explanation:

Radius of the pollen is given as

$r = 12.0 \mu m$

Volume of the pollen is given as

$V = \frac{4}{3}\pi r^3$

$V = \frac{4}{3}\pi (12\mu m)^3$

$V = 7.24 \times 10^{-15} m^3$

mass of the pollen is given as

$m = \rho V$

$m = 7.24 \times 10^{-12}$

so weight of the pollen is given as

$W = mg$

$W = (7.24 \times 10^{-12})(9.81)$

$W = 7.1 \times 10^{-11}$

Now electric force on the pollen is given

$F = qE$

$F = (-0.700\times 10^{-15})(95)$

$F = 6.65 \times 10^{-14} N$

now ratio of electric force and weight is given as

$\frac{F}{W} = \frac{6.65 \times 10^{-14}}{7.1 \times 10^{-11}}$

$\frac{F}{W} = 9.37 \times 10^{-4}$

7 0

3 years ago

Two ways that weak nuclear forces and electromagnetic forces are simaliar

STALIN [3.7K]

Have very short ranges

6 0

3 years ago