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3 years ago

Which statement about subatomic particles is NOT true?

Physics

1 answer:

san4es73 [151]3 years ago

4 0

A). Yes they do.
b). Yes they do.
c). Yes they don't.
d). No. See c).

So the answer is 'd'.

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6. An aircraft is is travelling along a runway at a Velocity of 25m/s in It accelerates at a rate of 4m/s² for a distance of 750

zysi [14]

Answer:

Take-off velocity = v = 81.39[m/s]

Explanation:

We can calculate the takeoff speed easily, using the following kinematic equation.

$v_{f}^{2}=v_{i}^{2} +2*a*x$

where:

a = acceleration = 4[m/s^2]

x = distance = 750[m]

vi = initial velocity = 25 [m/s]

vf = final velocity

$v_{f}=\sqrt{(25)^{2}+(2*4*750) } \\v_{f}=81.39[m/s]$

7 0

2 years ago

Options: A.) 10 N B.) 15 N C.) 25 N D.) 35N

Lady_Fox [76]

Given:

F_gravity = 10 N

F_tension = 25 N

Let's find the net centripetal force exterted on the ball.

Apply the formula:

$\sum ^{}_{}F_{\text{net}}=F_1+F_2=F_{centripetal}$

From the given figure, the force acting towards the circular path will be positive, while the force which points directly away from the center is negative.

Hence, the tensional force is positive while the gravitational force is negative.

Thus, we have:

$F_{\text{net}}=F_{\text{centripetal}}=F_{tension}-F_{gravity}=25N-10N=15N$

Therefore, the net centripetal force exterted on the ball is 15 N.

ANSWER:

15 N

7 0

8 months ago

A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carouse

Sidana [21]

Answer:

m = 35.98 Kg ≈ 36 Kg

Explanation:

I₀ = 125 kg·m²

R₁ = 1.50 m

ωi = 0.600 rad/s

R₂ = 0.905 m

ωf = 0.800 rad/s

m = ?

We can apply The law of conservation of angular momentum as follows:

Linitial = Lfinal

⇒ Ii*ωi = If*ωf <em>(I)</em>

where

Ii = I₀ + m*R₁² = 125 + m*(1.50)² = 125 + 2.25*m

If = I₀ + m*R₂² = 125 + m*(0.905)² = 125 + 0.819025*m

Now, we using the equation <em>(I) </em>we have

(125 + 2.25*m)*0.600 = (125 + 0.819025*m)*0.800

⇒ m = 35.98 Kg ≈ 36 Kg

5 0

3 years ago

Hi i need help. thNKS

Ede4ka [16]

Answer:

Explanation:

Groups go down, periods go across :)

3 0

2 years ago

An 805-kg race car can drive around an unbanked turn at a maximum speed of 54 m/s without slipping. the turn has a radius of cur

SIZIF [17.4K]

(a) Equating centripetal force to friction force, one finds the relation
v² = kar
for car speed v, coefficient of friction k, radius of curvature r, and downward acceleration a.

There is already downward acceleration due to gravity. The additional accceleration due to the wing is
a = F/m = 10600 N/(805 kg) ≈ 13.1677 m/s²
We presume this is added to the 9.80 m/s² gravity provides, so the coefficient of friction is
k = v²/(ar) = (54 m/s)²/((13.1677 m/s² +9.80 m/s²)·(155 m))
k ≈ 0.8191

(b) The maximum speed is proportional to the square root of the downward acceleration. Changing that by a factor of 9.80/(9.80+13.17) changes the maximum speed by the square root of this factor.
max speed with no wing effect = (54 m/s)√(9.8/22.97) ≈ 35.27 m/s

8 0

2 years ago