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3 years ago

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

Physics

2 answers:

VladimirAG [237]3 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]3 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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Hole filling fasteners (for example, MS20470 rivets) should not be used in composite structures primarily because of the?

DIA [1.3K]

Answer:

We can cause delamination.

Explanation:

The reason why is because the probability of causing delamination increase considerably when we use Hole-filling fasteners. If we use a typical rivet, these tends to expands in order to fill the hole.

If we analyze the force applied by the expanded rod will cause that the matrial will be deteriorated and will cause that the material to delaminate around the edges of the hole and we can cause possible control and no protection to the material.

7 0

3 years ago

Two hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 18.0 m . Sim

Nonamiya [84]

Answer:

The distance covered by puck A before collision is $z = 8.56 \ m$

Explanation:

From the question we are told that

The label on the two hockey pucks is A and B

The distance between the two hockey pucks is D 18.0 m

The speed of puck A is $v_A = 3.90 \ m/s$

The speed of puck B is $v_B = 4.30 \ m/s$

The distance covered by puck A is mathematically represented as

$z = v_A * t$

=> $t = \frac{z}{v_A}$

The distance covered by puck B is mathematically represented as

$18 - z = v_B * t$

=> $t = \frac{18 - z}{v_B}$

Since the time take before collision is the same

$\frac{18 - z}{V_B} = \frac{z}{v_A}$

substituting values

$\frac{18 -z }{4.3} = \frac{z}{3.90}$

=> $70.2 - 3.90 z = 4.3 z$

=> $z = 8.56 \ m$

8 0

3 years ago

At time t=0 a positively charged particle of mass m=3.57 g and charge q=9.12 µC is injected into the region of the uniform magne

larisa [96]

Answer:

10.78 s

Explanation:

The force on the charge is computed by using the equation:

$F^{\to}= qE^{\to} +q (v^{\to} + B^{\to}) \\ \\ F^{\to} = (9.12 \times 10^{-6}) *278 (-\hat k) +9.12 *10^{-6} *2.1 *0.18 (\hat i * \hat k) \\ \\ F^{\to} = -2.535 *10^{-3} \hat k -3.447*10^{-6} \hat j$

F = ma

∴

$a ^{\to}= \dfrac{F^{\to}}{m}$

$a ^{\to}= \dfrac{-1}{3.57\times 10^{-3}}(2.535*10^{-3}\hat k + 3.447*10^{-6} \hat j)$

$a ^{\to}=-0.710 \hat k -9.656*10^{-4} \hat j$

At time t(sec; the partiCle velocity becomes $v(t) = 3.78 v_o$

The velocity of the charge after the time t(sec) is expressed by using the formula:

$v^{\to}= v_{o \ \hat i} + a^{\to }t \\ \\ \implies (2.1)\hat i -0.710 t \hat k -9.656 \times 10^{-4} t \hat j = 3.78 v_o \\ \\ \implies (2.1)^2 +(0.710\ t)^2+ (9.656 *10^{-4}t )^2 = (3.78 *2.1^2 \\ \\ \implies 4.41 +0.5041 t^2 +9.324*10^{-7} t^2 = 63.012 \\ \\ \implies 4.41 +0.5041 t^2 = 63.012\\ \\ 0.5041t^2 = 63.012-4.41 \\ \\ t^2 = \dfrac{58.602}{0.5041} \\ \\ t^2 = 116.25 \\ \\ t = \sqrt{116.25} \\ \\ \mathbf{t = 10.78 \ s}$