How many protons would the element with the atomic number 10 contain?

Car A starts out traveling at 35.0 km/h and accelerates at 25.0 km/h2 for 15.0 min. Car B starts out traveling at 45.0 km/h and

lawyer [7]

1 kilometre=1000 metre

1 hour = 3600 second

$1\ km/hr=\frac{1000}{3600} m/s$

$1\ km/hr=\frac{5}{18} m/s$

The initial velocity of car A is 35.0 km/hr i.e

$35.0\ km/hr=35*\frac{5}{18} m/s$

= 9.72 m/s

The initial velocity of car B is 45 km/hr =12.5 m/s

The initial velocity of car C is 32 km/hr = 8.89 m/s

The initial velocity of car D is 110 km/hr=30.56 m/s

The acceleration of car A is given as $25\ km/hr^2$

$=\ 25*\frac{1000}{3600*3600} m/s^2$

$=0.00192901234 m/s^2$

The time taken by car A = 15 min.

From equation of kinematics we know that-

v= u+at [Here v is the final velocity and a is the acceleration and t is the time]

Final velocity of A, v = 9.72 m/s +[0.00192901234×15×60]m/s

=11.456111106 m/s

The acceleration of B is given as $15\ km/hr^2$

$=0.00115740740740 m/s^2$

The time taken by car B =20 min

The final velocity of B is -

v= u+at

= u-at [Here a is negative due to deceleration]

=12.5 m/s +[0.0011574074074×20×60]

=13.8888888.....

=13.9

The acceleration of C is given as $40\ km/hr^2$

$=\ 0.003086419753 m/s^2$

The time taken by car C =30 min

The final velocity of C is-

v = u+at

=8.89 m/s+[0.003086419753×30×60] m/s

=14.4455555555..m/s

=14.45 m/s

The car C is decelerating.The deceleration is given as- $60\ km/hr^2$

$=0.0046296296296m/s^2$

The time taken by car D= 45 min.

The final velocity of the car D is-

v =u+at

=30.56 -[0.00462962962962×45×60]m/s

=18.06 m/s

Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.

3 0

3 years ago

A image formed in a plane mirror is larger than the object ture or false

Lesechka [4]

Answer:

false

Explanation:

8 0

2 years ago

At what height h above the ground does the projectile have a speed of 0.5v?

dedylja [7]

Answer:height above ground at which projectile have velocity

0.5v is (0.0375v^2)

Explanation:

Using Vf = Vi - gt

Where Vf is final velocity

Vi is initial velocity

g is the acceleration due to gravity

t is the time taken

So, 0.5v = v - gt

t = 0.05v

Therefore height h = vt - 0.5gt^2

Subtitute t

h = 0.05v^2 - 0.0125v^2

h = 0.0375v^2

3 0

2 years ago

A blue line with 5 orange tick marks then one red tick mark then 4 orange tick marks. The number zero is above the red tick mark

Digiron [165]

Answer:

16 cm

Explanation:

Given that,

The object begins from 0 and moves 3cm towards left side followed by 7 cm towards the right and then, 6 cm towards the left side.

Let the x-axis to be the +ve and on the right side and -ve on the left

Thus, displacement would be:

= 0 -3 + 7 -6

= -2 cm

This implies that the object displaces 2cm towards the left.

While the total distance covered by the object equal to,

= 0cm + 3cm + 7cm + 6cm

= 16 cm

Thus, 16 cm is the total distance.

3 0

3 years ago

A gray kangaroo can bound across a flat stretch of ground with each jump carrying it 11 m from the takeoff point. part a if the

Svetllana [295]

We have that the maximum rank of the kangaroo is given by:
R = v0 ^ 2 sin (2θ) / g
where,
v0 = initial velocity
θ = angle of the velocity vector formed from the horizontal
g = gravity
Clearing the speed we have:
v0 ^ 2 = (R * g) / (sin (2θ))
Substituting values
v0 = root (((11) * (9.8)) / (sin (2 (21 * (pi / 180)))))
v0 = 12.69 m / s
answer
its takeoff speed is 12.69 m / s

6 0

3 years ago

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