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2 years ago

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Baseballs pitched by a machine have a horizontal velocity of 30 meters/second. The machine accelerates the baseball from 0 meter

s/second to 30 meters/second in 0.5 seconds. If a baseball has a mass of 0.15 kilograms, the force the machine exerts is _______ newtons. Use F = ma, where a= v-u/t.

1 answer:

miv72 [106K]2 years ago

4 0

using the formula

F = ma

Where F is the force applied by the machine

A is the acceleration which is also equal to v/t where v is the velocity and t is time

M is the mass

F = mv/t

F = (0.15kg) (30 – 0 m/s)/ 0.5 s

<span>F = 9 N</span>

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Find the given attachment for solution

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2 years ago

1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n

sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

$(1).\: x =v_0t$

$(2).\: y= 15m-\dfrac{1}{2}gt^2$

where $v_0$ is the initial velocity.

(a).

When the projectile hits the 50m mark, $y=0$ ; therefore,

$0=15-\dfrac{1}{2}gt^2$

solving for $t$ we get:

$t= 1.75s.$

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

$50m = v_0(1.75s)$

which gives

$\boxed{v_0 = 28.58m/s.}$

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

$v_x = 28.58m/s.$

the vertical component of the velocity is

$v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.$

which gives a speed $v$ of

$v = \sqrt{v_x^2+v_y^2}$

$\boxed{v =33.3m/s.}$

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1 year ago

Hippos spend much of their lives in water, but amazingly, they don’t swim. manatees, They have, like little very body fat. The d

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Answer:

428.59 N

Explanation:

Buoyant force, $B=Vg\rho$ where V is volume, g is gravitational constant and \rho is density

$B+F_{upward}=mg$ where $F_{upward}$ is upward force

$Vg\rho_{w}+F_{upward}=mg$

$F_{upward}=mg- Vg\rho_{w}$

$F_{upward}=g(mg- V\rho_{w})=g(m-m\frac {\rho_{w}{\rho_{hippo}}$ where $\rho_{hippo}$ is the density of hippo

$F_{upward}=mg(1-\frac {\rho_{w}}{\rho_{hippo}})$

Using g as 9.81

$F_{upward}=1500*9.81*(1-1000/1030)= 428.5922 N$

Therefore, the upward force=428.59 N

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2 years ago

16) A wheel of moment of inertia of 5.00 kg-m2 starts from rest and accelerates under a constant torque of 3.00 N-m for 8.00 s.

KiRa [710]

Answer:

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Rotational kinetic energy of a body can be determined using the expression

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I is the moment of inertia around axis of rotation. = 5kgm/s²

ω is the angular velocity = ?

Note that torque (T) = I¶ where;

¶ is the angular acceleration.

I is the moment of inertia

¶ = T/I

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Angular acceleration (¶) = ∆ω/∆t

∆ω = ¶∆t

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Therefore, rotational kinetic energy = 1/2×5×4.8²

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2 years ago

Read 2 more answers

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Annette [7]

Jogger moves in three displacements

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d2 = 5 blocks South

d3 = 2 blocks East

now we can say

total displacement towards East direction will be

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Total displacement towards South

$d_y = 5 block$

now to find the net displacement we can use vector addition

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$d = \sqrt{12^2 + 5^2}$

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<em>so magnitude of net displacement will be equal to 13 blocks</em>

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2 years ago