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3 years ago

A car starts from rest and accelerates for 5.2 s

Physics

1 answer:

vodomira [7]3 years ago

6 0

Answer:

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Explanation:

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Cobalt (Co) and nickel (Ni) have almost identical atomic mass numbers. Comparing the typical nickel and cobalt atoms, which stat

Ivahew [28]

D. Cobalt has one less proton.

Explanation:

Comparing a typical nickel and cobalt atom suggests that cobalt has one less proton in its nuclear space.

Cobalt has an atomic number of 27

Nickel has an atomic number of 28

Cobalt and nickel are both transition metals found in the d-block on the periodic table.

Atomic number is the number of protons in the nucleus of an atom.

We can see that Nickel has 1 more proton compared to cobalt.

Learn more:

Atomic number brainly.com/question/2057656

#learnwithBrainly

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4 years ago

Read 2 more answers

A 2.0 kg wood block is launched up a wooden ramp that is inclinedat a 35* angle. The block’s initial speed is 10m/s. (Giventhat

IgorC [24]

a) 4.0 m

We can solve this part by writing the equations of motion along the two directions: perpendicular to the slope and parallel to the slope.

Perpendicular to the slope:

$N-mgcos \theta =0$ (1)

where N is the normal reaction, m = 2.0 kg is the mass of the block, g = 9.8 m/s^2 is the acceleration of gravity, $\theta=35^{\circ}$ is the angle.

Parallel to the slope:

$-\mu_k N -mgsin \theta = ma$ (2)

where $\mu_k=0.20$ is the coefficient of friction, and a the acceleration, and where we have chosen up the slope as positive direction, so both forces are negative.

From (1) we get

$N=mg cos \theta$

And substituting into (2), we can find the acceleration:

$-\mu_k mg cos \theta -mgsin \theta = ma\\a=-\mu_k g cos\theta - g sin \theta = -(0.20)(9.8)(cos 35^{\circ})-(9.8)(sin 35^{\circ})=-7.2 m/s^2$

where the negative sign means the direction is down the slope.

Now we can find the distance travelled along the slope by using the SUVAT equation

$v^2-u^2=2ad$

where

v = 0 is the velocity when the block comes to rest

u = 10 m/s is the initial velocity

d is the distance travelled along the slope

Solving for d,

$d=\frac{v^2-u^2}{2a}=\frac{0-(10)^2}{2(-7.2)}=6.9 m$

And so, the vertical heigth gained by the block is

$h=d sin \theta = (6.9)(sin 35^{\circ})=4.0 m$

b) 7.4 m/s

The equation of motion along the direction parallel to the slope in this case is

$-\mu_k N +mgsin \theta = ma$

where this time we have taken down the slope as positive direction, so the component of the weight is positive while the frictional force is negative since the block slides downward (while friction acts upward). Solving for a, we find the new acceleration:

$a=-\mu_k g cos \theta + g sin \theta = -(0.20)(9.8)(cos 35^{\circ})+(9.8)(sin 35^{\circ})=4.0 m/s^2$