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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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A mass M of 3.80E-1 kg slides inside a hoop of radius R=1.10 m with negligible friction. When M is at the top, it has a speed of

Vikentia [17]

Answer:

N = 26.59 N

Explanation:

given,

mass = 0.38 kg

radius of the hoop = 1.10 m

speed = 5.35 m/s

force = ?

now,

$\dfrac{1}{2}mv_t^2 + mg(2R) = \dfrac{1}{2}mv^2 + mgR(1-cos \theta)$

$mv^2 = mv_t^2 + 2mgR(1 + cos \theta)$

we know that,

$N - mgcos \theta = \dfrac{mv^2}{R}$

$N - mgcos \theta = \dfrac{mv_t^2 + 2mgR(1 + cos \theta)}{R}$

$N - mgcos \theta = \dfrac{mv_t^2 }{R}+ 2mg(1 + cos \theta)$

$N = \dfrac{mv_t^2 }{R}+ 2mg + 3mgcos \theta)$

$N = \dfrac{0.38\times 5.35^2 }{1.1}+ 2\times 0.38\times 9.8 + 3\times 0.38 \times 9.8 cos 34^0)$

N = 26.59 N

3 0

3 years ago

A 0.046 kg golf ball hit by a driver can accelerate from rest to 67 m/s in 1 ms while the driver is in contact with the ball. Ho

Flauer [41]

Answer:

Average force = 67 mn

Explanation:

Given:

Initial velocity u = 0 m/s

Final velocity v = 67 m/s

Time t = 1 ms = 0.001 sec.

Computation:

Using Momentum theory

Change in momentum = F × Δt

(v-u)/t = F × Δt

F × 0.001 = (67 - 0)/0.001

F= 67,000,000

Average force = 67 mn

3 0

2 years ago

Which substance is a combination of different atoms?

defon

Every chemical "compound" is. Some examples of compounds include ...

Salt
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7 0

3 years ago

Whats 80.000g = what kg

Rama09 [41]

there is 1000 grams in 1 kilogram

Divide

80/1000 = 0.08

0.08 kilograms is your answer

hope this helps

3 0

3 years ago

The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This

erica [24]

To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

$g = \frac{GM}{(d-R_{CM})^2}$

Where,

G = Gravitational Universal Constant

d = Distance

M = Mass

$R_{CM} =$ Radius earth center of mass

PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,

$g = \frac{GM}{(d-R_{CM})^2}$

$g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}$

$g = 3.4*10^{-5}m/s^2$

PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

$\omega = \frac{2\pi}{T}$

At the same time we have that centripetal acceleration is given as

$a_c = \omega^2 r$

Replacing

$a_c = (\frac{2\pi}{T})^2 r$

$a_c = (\frac{2\pi}{26.3d(\frac{86400s}{1days})})^2 (4700*10^3m)$

$a_c = 3.34*10^{-5}m/s^2$

3 0

3 years ago