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2 years ago

Question 2 of 10

Physics

1 answer:

Anastaziya [24]2 years ago

5 0

Answer:

Explanation:

The question does not specify any outside forces that could slow down the ball horizontally. There fore the ball does not accelerate or decelerate horizontally. Therefore, a = 0m/s2

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A 15 kg dog jumps out a stationary sled which has a mass of 40 kg. If

zzz [600]

Velocity of the sled is 3.2 m/s

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2 years ago

The formula v = √2.5r models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radiu

mel-nik [20]

Answer:

The maximum safe speed of the car is 30.82 m/s.

Explanation:

It is given that,

The formula that models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radius of curvature r r, is in feet is given by :

$v=\sqrt{2.5\ r}$ .........(1)

A highway crew measures the radius of curvature at an exit ramp on a highway as 380 feet, r = 380 feet

Put the value of r in equation (1) as :

$v=\sqrt{2.5\ \times 380}$

v = 30.82 m/s

So, the maximum safe speed of the car is 30.82 m/s. Hence, this is the required solution.

4 0

3 years ago

The uncertainty in the position of an electron along an x axis is given as 5 x 10-12 m. What is the least uncertainty in any sim

Vsevolod [243]

Answer:

The least uncertainty in the momentum component px is 1 × 10⁻²³ kg.m.s⁻¹.

Explanation:

According to Heisenberg's uncertainty principle, the uncertainty in the position of an electron (σx) and the uncertainty in its linear momentum (σpx) are complementary variables and are related through the following expression.

σx . σpx ≥ h/4π

where,

h is the Planck´s constant

If σx = 5 × 10⁻¹²m,

5 × 10⁻¹²m . σpx ≥ 6.63 × 10⁻³⁴ kg.m².s⁻¹/4π

σpx ≥ 1 × 10⁻²³ kg.m.s⁻¹

4 0

3 years ago

Inertia is ______ to mass.

UkoKoshka [18]

Answer:

Inertia is directly proportional to mass of an object. Therefore, when the force of inertia increases the mass also increases, and when it decreases the mass also decreases.

Explanation:

6 0

3 years ago

Read 2 more answers

An electron moving parallel to a uniform electric field increases its speed from 2.0 × 10^7 m/s to 4.0 × 10^7 m/s over a distanc

julia-pushkina [17]

Answer:

$E = 2.84 * 10^5 N/C$

Explanation:

The speed increased from 2.0 * 10^7 m/s to 4.0 * 10^7 m/s over a 1.2 cm distance.

Let us find the acceleration:

$v^2 = u^2 + 2as$

$(4.0 * 10^7)^2 = (2.0 * 10^7)^2 + 2 * a * 0.012\\\\(4.0 * 10^7)^2 - (2.0 * 10^7)^2 = 0.024a\\\\1.2 * 10^{15}= 0.024a\\\\a = 1.2 * 10^{15} / 0.024\\\\a = 5 * 10^{16} m/s^2$

Electric force is given as the product of charge and electric field strength:

F = qE

where q = electric charge

E = Electric field strength

Force is generally given as:

F = ma

where m = mass

a = acceleration

Equating both:

ma = qE

E = ma / q

For an electron:

m = 9.11 × 10^{-31} kg

q = 1.602 × 10^{-19} C

Therefore, the electric field strength of the electron is:

$E = \frac{9.11 * 10^{-31} * 5 * 10^{16}}{1.602 * 10^{-19}} \\\\E = 2.84 * 10^5 N/C$

8 0

3 years ago