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2 years ago

A football is thrown at an angle of 30.° above the horizontal. The magnitude of the horizontal

Physics

2 answers:

VladimirAG [237]2 years ago

8 0

Answer:

1.53 s

Explanation:

Initially vertical component of velocity of the ball, uy = 7.5 m/s

Net displacement is vertical direction is zero, Δy =0

Use second equation of motion:

Δy = uy t + 0.5 a t²

Here, acceleration a = -g (g =9.8 m/s²)

Substitute all the values and solve for g

0 = 7.5 t -0.5 (9.8)t²

7.5 t = 4.9 t²

t = 1.53 s

just olya [345]2 years ago

4 0

Answer:

Explanation:

Let the initial velocity is u.

ux = 13 m/s

uy = 7.5 m/s

Angle of projection, θ = 30°

ux = u Cos 30 = 13

u = 15 m/s

Height caught is same

use second equation of motion

h = uy t - 0.5 x gt²

0 = 75. t - 0.5 x 9.8 x t²

t = 1.24 s

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When a gas is heated, it absorbs 196 joules of heat from the surroundings. At the same time, the gas expands, doing pressure-vol

timurjin [86]

Answer:

(a) q positive; w negative.

(b) ΔE = -126 J

Explanation:

(a) Determine whether the amounts of heat (q) and work (w) exchanged should have positive or negative signs. heat (q) positive negative work (w) positive negative

By convention, when the system absorbs heat from the surroundings, its sign is positive, that is, q = 196 J.

By convention, when the system exerts work on the surroundings, its sign is negative, that is, w = -322 J.

(b) Calculate the change in internal energy (ΔE) of the gas. J

The change in internal energy (ΔE) can be calculated using the following expression.

ΔE = q + w

ΔE = 196 J + (-322 J) = -126 J

(c) Determine whether one or more of the following is a state function:

internal energy (E) of a system,

change in internal energy (ΔE) of a system,

heat (q) absorbed or released by a system,

work (w) done on or by a system.

E and ΔE are state functions (they only depend on the states of the gas), whereas q and w depend on the trajectory.

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2 years ago

What is a list of solid mixtures in a classroom?

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3 years ago

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Vera_Pavlovna [14]

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2 years ago

A 12.0-g bullet is fired horizontally into a 104-g wooden block that is initially at rest on a frictionless horizontal surface a

kobusy [5.1K]

Answer:

v₀ = 292.3 m / s

Explanation:

Let's analyze the situation, on the one hand we have the shock between the bullet and the block that we can work with at the moment and another part where the assembly (bullet + block) compresses a spring, which we can work with mechanical energy, as the data they give us are Let's start with this second part.

We write the mechanical energy when the shock has passed the bodies

Em₀ = K = ½ (m + M) v²

We write the mechanical energy when the spring is in maximum compression

$Em_{f}$ = $K_{e}$ = ½ k x²

Em₀ = $Em_{f}$

½ (m + M) v² = ½ k x²

Let's calculate the system speed

v = √ [k x² / (m + M)]

v = √[154 0.83² / (0.012 +0.104) ]

v = 30.24 m / s

This is the speed of the bullet + Block system

Now let's use the moment to solve the shock

Before the crash

p₀ = m v₀

After the crash

$p_{f}$ = (m + M) v

The system is formed by the bullet and block assembly, so the forces during the crash are internal and the moment is preserved

p₀ = $p_{f}$

m v₀ = (m + M) v

v₀ = v (m + M) / m

let's calculate

v₀ = 30.24 (0.012 +0.104) /0.012

v₀ = 292.3 m / s

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2 years ago