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2 years ago

What is horizontal motion

Physics

2 answers:

Stella [2.4K]2 years ago

6 0

Answer:

A projectile moves along its path with a constant horizontal velocity

Klio2033 [76]2 years ago

4 0

Horizontal motion is defined as a projectile motion in a horizontal plane depending upon the force acting on it. For a short distance, the vertical and horizontal components of a projectile are perpendicular and independent of each other.

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Acceleration is measured in_________ m g m/s m/s2

Nana76 [90]

Acceleration is measured in m/s².

Answer: m/s²

4 0

3 years ago

An electron and a 0.033 0-kg bullet each have a velocity of magnitude 495 m/s, accurate to within 0.010 0%. Within what lower li

lara31 [8.8K]

Answer:

1.170*10^-3 m

3.23*10^-32 m

Explanation:

To solve this, we apply Heisenberg's uncertainty principle.

the principle states that, "if we know everything about where a particle is located, then we know nothing about its momentum, and vice versa." it also can be interpreted as "if the uncertainty of the position is small, then the uncertainty of the momentum is large, and vice versa"

Δp * Δx = h/4π

m(e).Δv * Δx = h/4π

If we make Δx the subject of formula, by rearranging, we have

Δx = h / 4π * m(e).Δv

on substituting the values, we have

for the electron

Δx = (6.63*10^-34) / 4 * 3.142 * 9.11*10^-31 * 4.95*10^-2

Δx = 6.63*10^-34 / 5.67*10^-31

Δx = 1.170*10^-3 m

for the bullet

Δx = (6.63*10^-34) / 4 * 3.142 * 0.033*10^-31 * 4.95*10^-2

Δx = 6.63*10^-34 / 0.021

Δx = 3.23*10^-32 m

therefore, we can say that the lower limits are 1.170*10^-3 m for the electron and 3.23*10^-32 for the bullet

7 0

3 years ago

Predict the deformation or elongation of a spring that has a constant of elasticity of 400 N/m when a force of 75 N is applied i

morpeh [17]

Answer:

Explanation:

Give that,

Spring constant (k)=40N/m

Force applied =75N

Since the force is applied to the right, we don't know if it is compressing or stretching the spring

So let assume it compress

Using hooke's law

F=-ke

e=-F/k

Then, e=-75/40

e=-1.875m

The deformation is 1.875m.

Let assume it stretch

Using hooke's law

-F=-ke

e=F/k

Then, e=75/40

e=1.875m

The elongation is 1.875m

3 0

3 years ago

As part

umka21 [38]

Answer:

Part a)

$a = 3.68 m/s^2$

Part b)

$a = 11.8 m/s^2$

Explanation:

Part a)

For force conditions of two blocks we will have

$m_1g - T = m_1 a$

$T - m_2g = m_2 a$

now from above equations we have

$(m_1 - m_2) g = (m_1 + m_2) a$

$a = \frac{m_1 - m_2}{m_1 + m_2} g$

now we know that

$m_1 = \frac{908}{9.8} = 92.65 kg$

$m_2 = \frac{412}{9.8} = 42 kg$

now from above equation we have

$a = \frac{92.65 - 42}{92.65 + 42}(9.8)$

$a = 3.68 m/s^2$

Part b)

When heavier block is removed and F = 908 N is applied at the end of the string then we have

$F - mg = ma$

$908 - 412 = 42 a$

$a = 11.8 m/s^2$

8 0

3 years ago

A spherical snowball melts at a rate proportional to its surface area. (a) write a differential equation for its volume, v. (use

tresset_1 [31]

Answer:

Explanation:

The rate of change in volume is proportional to the surface area:

dV/dt = kA

Integrating:

V = kAt + C

At t=0, V = s, so:

s = kA(0) + C

C = s

Therefore:

V = kAt + s

6 0

3 years ago