Find the speed for Barkely, the dog, if he runs 5.5 miles in 3 hours

The ability to pick up small objects, to stack objects, and to carry objects appears first during which time period?

aliya0001 [1]

the answer is B 4-6 months thats when they can pick up large objects.

8 0

2 years ago

A body travels at an initial speed of 2.5 m/s. Given a constant acceleration of 0.2 m/s 2 what is the speed of the body at time

garri49 [273]

Answer:

<u>We are given:</u>

u = 2.5 m/s

a = 0.2 m/s/s

t = 25 seconds

v = v m/s

<u>Solving for 'v':</u>

From the first equation of motion:

v = u + at

Replacing the values

v = 2.5 + (0.2)(25)

v = 2.5 + 5

v = 7.5 m/s

6 0

3 years ago

a pendulum, 2.0 m in length, is released with a push when the string is at an angle of 25 o from the vertical. if the initial sp

stiks02 [169]

Answer:

1 / 2 m v^2 = L m g (1 - cos θ)

This is the KE due to the pendulum falling from a 25 deg displacement

v^2 = 2 L g (1 - cos 25) = 2 * 2 * 9.8 (1 - .906) = 3.67 m^2/s^2

v = 1.92 m/s this is the speed due to an initial displacement of 25 deg

Its speed at the bottom would then be

1.92 + 1.2 = 3.12 m/s since it gains 1.92 m/s from its initial displacement

3 0

1 year ago

One hundred turns of insulated copper wire are wrapped into a circular coil of crosssectional area 1.20⇥103 m2. The two ends of

arsen [322]

Answer:

236.3 x $10^-^3$ C

Explanation:

Given:

B(0)=1.60T and B(t)=-1.60T

No. of turns 'N' =100

cross-sectional area 'A'= 1.2 x $10^-^3$ m²

Resistance 'R'= 1.3Ω

According to Faraday's law, the induced emf is given by,

ℰ=-NdΦ/dt

The current given by resistance and induced emf as

I = ℰ/R

I= -NdΦ/dtR

By converting the current to differential form(the time derivative of charge), we get

$\frac{dq}{dt}$ = -NdΦ/dtR

dq= -N dΦ/R

The change in the flux dФ =Ф(t)-Ф(0)

therefore, dq = $\frac{N}{R}$ (Ф(0)-Ф(t))

Also, flux is equal to the magnetic field multiplied with the area of the coil

dq = NA(B(0)-B(t))/R

dq= (100)(1.2 x $10^-^3$ )(1.6+1.6)/1.3

dq= 236.3 x $10^-^3$ C

5 0

3 years ago

Objects with masses of 135 kg and a 435 kg are separated by 0.500 m. Find the net gravitational force exerted by these objects o

IrinaVladis [17]

Answer:

The net gravitational force on the mass is $1.27\times 10^{-5}$

Explanation:

We have by Newton's law of gravity the force of attraction between masses $m_{1},m_{2}$

$F_{att}=G\frac{m_{1}m_{2}}{r^{2}}$

Applying vales we get

Force of attraction between 135 kg mass and 38 kg mass is

$F_{1}=6.67\times 10^{-11}\frac{135\times 38}{(0.25)^{2}}\\\\F_{1}=5.47\times 10^{-6}N$

Force of attraction between 435 kg mass and 38 kg mass is

$F_{2}=6.67\times 10^{-11}\frac{435\times 38}{(0.25)^{2}}\\\\F_{2}=1.76\times 10^{-5}N$

Thus the net force on mass 38.0 kg is $F_{2}-F_{1}=1.76\times 10^{-5}-5.47\times 10^{-6}\\\\F_{2}-F_{1}=1.27\times 10^{-5}$

8 0

3 years ago