All categories

2 years ago

After a projectile is fired into the air, what is the acceleration in the x-direction? (Assume no air resistance.)

Physics

2 answers:

storchak [24]2 years ago

5 0

Answer:

$a_x = 0$

Explanation:

When projectile is in air then the force on the projectile is only due to its weight

It is vertically downwards

$F_g = mg(-\hat j)$

so here since we don't have any air drag or friction due to air so there will be no force along its tangent or along x direction

so the acceleration along x direction will be zero

$a_x = 0$

$a_y = - g$

Gre4nikov [31]2 years ago

4 0

Without air resistance, there is no horizontal force on the projectile,
so it has no horizontal acceleration. The answer is "zero".

You might be interested in

A student does 25 J of work on the handle of a pencil sharpener. If the pencil sharpener does 20 J of work on the pencil, what i

allsm [11]

Answer:

80%

Explanation:

Efficiency of machine = work output/work input ×100 %

From question, work output = 20J

Work input = 25J.

Therefore efficiency = 20/25 × 100 %

Efficiency = 20×4 %

Efficiency = 80%

I hope this was helpful, please mark as brainliest

3 0

2 years ago

A 49.5 gram super ball traveling at 25.5 m/s bounces off a brick wall and rebounds at 19.5 m/s. a high-speed camera records this

OLga [1]

T = 4.25 ms = 4 x 10⁻³ s, the time for rebound
v₁ = 25.5 m/s, the impacting velocty
v₂ = -19.5 m/s, the rebounding velocity (n the opposite directon)

The change in velocity is
v₂ - v₁ = - (25.5+19.5) = -45 m/s

The acceleration is
a = (-45 m/s)/(4 s) = -11.25 m/s²
The negative sign indicates that the final velocity is opposiye to the impact velocty.

Answer: The magnitude of the acceleration is 11.25 m/s²

8 0

2 years ago

We are designing a crude propulsion mechanism for a science fair demonstration. One of our team members stands on a skateboardth

Scrat [10]

Answer:

greater speed will be obtained for the elastic collision,

Explanation:

To answer this exercise we must find the speed that the sail acquires after each impact.

Let's start by hitting a ball of clay.

The system is formed by the candle and the clay balls, therefore the forces during the collision are internal and the moment is conserved.

initial instant. before the crash

p₀ = m v₀

where m is the mass of the ball and vo its initial velocity, we are assuming that the candle is at rest

final instant. After the crash

the mass of the candle is M

p_f = (m + M) v

the moment is preserved

p₀ = p_f

m v₀ = (m + M) v

v = $\frac{m}{m+M} \ v_o$

for when n balls have collided

v = $\frac{m}{n \ m + M}$ v₀

Now let's analyze the case of the bouncing ball (elastic)

initial instant

p₀ = m v₀

final moment

p_f = m v_{1f} + M v_{2f}

p₀ = p_f

m v₀ = m v_{1f} + M v_{2f}

m (v₀ - v_{1f}) = M v_{2f}

this case corresponds to an elastic collision whereby the kinetic energy is conserved

K₀ = K_f

½ m v₀² = ½ m v_{1f}² + ½ M v_{2f}²

v₁ = v_{1f} v₂ = v_{2f}

m (v₀² - v₁²) = M v₂²

let's use the identity

(a² - b²) = (a + b) (a-b)

we write our equations

m (v₀ - v₁) = M v₂ (1)

m (v₀ - v₁) (v₀ + v₁) = M v₂²

let's divide these equations

v₀ + v₁ = v₂

Let's look for the final speeds

we substitute in equation 1

m (v₀ - v₁) = M (v₀ + v₁)

v₀ (m -M) = (m + M) v₁

v₁ = $\frac{m-M}{m + M}$ v₀

we substitute in equation 1 to find v₂

$\frac{M}{m}$ v₂ = v₀ - $\frac{m-M}{m+M}$ v₀

v₂ = $\frac{m}{M} ( 1 - \frac{m-M}{m+M} ) \ v_o$

v₂ = $\frac{m}{M} ( \frac{2M}{m+M} ) \ \ v_o$

v₂ = $\frac{2m}{m +M} \ v_o$

Let's analyze the results for inelastic collision with each ball that collides with the sail, the total mass becomes larger so the speed increase is smaller and smaller.

In the case of elastic collision, the increase in speed is constant with each ball since the total mass remains invariant.

Consequently, greater speed will be obtained for the elastic collision, that is, the ball will bounce.

8 0

2 years ago

a load of 45 N attached to a spring that is hanging vertically stretches the spring 0.14 m. What is the spring constant?

zhenek [66]

According to Hooke's law, Force = spring constant x displacement of the spring. Spring constant = Force/displacement in spring = 45/0.14 = 321.42 N/m. Hope this helps!

3 0

2 years ago

Read 2 more answers

Competitive pressures stemming from the threat of entry are stronger when

navik [9.2K]

Answer:

Industries outlook is uncertain

Explanation:

Competitive pressures stemming from the threat of entry are stronger when the industry's outlook is uncertain or highly risky, entry barriers are low, and very few existing industry members are looking to expand their market reach by entering product segments or geographic areas where they currently do not have a presence. entry barriers are low, the pool of entry candidates is large, and existing industry members are earning good profits. there are fewer than 10 entry candidates with the potential to hurdle the industry's barriers to entry. t is difficult or costly for a customer to switch to a new brand, the total dollar investment needed to enter the market successfully exceeds $5 million, and existing governmental regulations impose significant cost and compliance burdens on industry members. buyers have strong brand preferences and high degrees of loyalty to their preferred brand and when it takes new entrants less than 5 years to secure attractive amounts of space on retailers' shelves and build a well-recognized brand name.

6 0

2 years ago