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jonny [76]
2 years ago
13

An object with 10 protons 14 electrons and 5 neutrons would have what charge

Physics
1 answer:
True [87]2 years ago
7 0

Answer:-4

Explanation:because

10protone +10

14 electrone-14

5 neutrones0 the result will be

10-14=-4

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A projectile is launched with an initial velocity of 25 m/s at an angle of 30° above the horizontal. The projectile reaches maxi
Alchen [17]

Answer:

x ≈ 56 m

Explanation:

vertical initial velocity =v_{0y}(t) = 25 m/s* sin(30°)= 12.5 m/s

height = h

h =v_{0y}t+\frac{at^{2}}{2} \\\\65m = 12.5m/s*t + \frac{9.8m/s^{2}*t^{2}}{2} \\\\t=2.584 s

t- time is found solving quadratic equation.

horizontal velocity = v_{0x}=25m/s*cos(30^{o})=21.65 m/s

Horizontal velocity is constant, so distance x=v_{0x}*t =21.65 m/s *2.584 s=55.9 = 56 m

6 0
3 years ago
A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. its initial position
Nesterboy [21]

The position of the object at time t =2.0 s is <u>6.4 m.</u>

Velocity vₓ of a body is the rate at which the position x of the object changes with time.

Therefore,

v_x= \frac{dx}{dt}

Write an equation for x.

dx=v_xdt\\ x=\int v_xdt

Substitute the equation for vₓ =2t² in the integral.

x=\int v_xdt\\ =\int2t^2dt\\ =\frac{2t^3}{3} +C

Here, the constant of integration is C and it is determined by applying initial conditions.

When t =0, x = 1. 1m

x= \frac{2t^3}{3} +C\\ x_0=1.1\\ x= (\frac{2t^3}{3} +1.1)m

Substitute 2.0s for t.

x= (\frac{2t^3}{3} +1.1)m\\ =\frac{2(2.0)^3}{3} +1.1\\ =6.43 m

The position of the particle at t =2.0 s is <u>6.4m</u>




5 0
2 years ago
Explain the difference between the terms "rotation" and "revolution"
Novay_Z [31]

Answer:

Explanation:

"Rotation" refers to an object's spinning motion about its own axis. "Revolution" refers the object's orbital motion around another object. For example, Earth rotates on its own axis, producing the 24-hour day. Earth revolves about the Sun, producing the 365-day year.

3 0
2 years ago
Read 2 more answers
A 5 kg ball rolling at 1.0 m/s hits a 15 kg ball at rest. The balls stick together after the collision What is the
Reika [66]

Answer:

0.25m/s

Explanation:

Given parameters

m₁  = 5kg

v₁ = 1.0m/s

m₂ = 15kg

v₂ = 0m/s

Unknown:

velocity after collision = ?

Solution:

Momentum before collision and after collision will be the same. For inelastic collision;

    m₁v₁ + m₂v₂  = v(m₁ + m₂)

Insert parameters and solve for v;

   5 x 1  + 15 x 0 = v (5 + 15 )

          5  = 20v

          v = \frac{5}{20}   = 0.25m/s

5 0
3 years ago
Consider a block on frictionless ice. Starting from rest, the block travels a distance din
sweet [91]

Answer:

<em>The distance is now 4d</em>

Explanation:

<u>Mechanical Force</u>

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = m.a

Where a is the acceleration of the object.

The acceleration can be calculated by solving for a:

\displaystyle a=\frac{F}{m}

Once we know the acceleration, we can calculate the distance traveled by the block as follows:

\displaystyle d = vo.t+\frac{at^2}{2}

If the block starts from rest, vo=0:

\displaystyle d = \frac{at^2}{2}

Substituting the value of the acceleration:

\displaystyle d = \frac{\frac{F}{m}t^2}{2}

Simplifying:

\displaystyle d = \frac{Ft^2}{2m}

When a force F'=4F is applied and assuming the mass is the same, the new acceleration is:

\displaystyle a'=\frac{4F}{m}

And the distance is now:

\displaystyle d' = \frac{4Ft^2}{2m}

Dividing d'/d:

\displaystyle \frac{d' }{d}=\frac{\frac{4Ft^2}{2m}}{\frac{Ft^2}{2m}}

Simplifying:

\displaystyle \frac{d' }{d}=4

Thus:

d' = 4d

The distance is now 4d

3 0
2 years ago
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