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2 years ago

12

b) When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 2.50 m/s

. What distance in meters does the ball travel horizontally

1 answer:

fomenos2 years ago

3 0

Answer:

The horizontal distance is 0.64 m.

Explanation:

Initial velocity, u =2.5m/s

The maximum horizontal distance is

$R = \frac{u^2}{2g}\\\\R = \frac{2.5\times 2.5}{9.8}\\\\R = 0.64 m$

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Compute the dot product of the vectors u and v, and find the angle between the vectors. Bold v equals 7 Bold i minus Bold j and

OLga [1]

Answer:

$\theta = 106.3 degree$

Explanation:

As we know that

$\vec w = -\hat i + 7\hat j$

$\vec v = 7\hat i - \hat j$

also we know that

$\vec v. \vec w = -14$

it is given as

$\vec v. \vec w = (-\hat i + 7\hat j).(7\hat i - \hat j)$

$\vec v. \vec w = - 7 - 7 = -14$

also we can find the magnitude of two vectors as

$|v| = \sqrt{(-1)^2 + (7)^2}$

$|v| = \sqrt{50}$

similarly we have

$|w| = \sqrt{(7^2) + (-1)^2}$

$|w| = \sqrt{50}$

now we know the formula of dot product as

$\vec v. \vec w = |v||w| cos\theta$

$-14 = (\sqrt{50})^2cos\theta$

$\theta = cos^{-1}(\frac{-14}{50})$

$\theta = 106.3 degree$

3 0

2 years ago

the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force

-Dominant- [34]

Answer:

PART A)

External force will be 75 N

PART B)

distance moved will be 1.125 m

Explanation:

PART A)

Given that net force on the mower is

$F_{net} = 51 N$

now we also know that friction force due to ground is given as

$F_f = 24 N$

now we have

$F_{net} = F_{ext} - F_f$

$51 = F_{ext} - 24$

$F_{ext} = 75 N$

so external force will be 75 N

PART B)

deceleration due to friction when external force is removed from it

$a = \frac{F_f}{m}$

$a = \frac{24}{24} = 1 m/s^2$

now we can find the distance by kinematics

$v_f^2 - v_i^2 = 2 a d$

$0 - 1.5^2 = 2(-1)d$

$d = 1.125 m$

so the distance moved will be 1.125 m

6 0

2 years ago

Consider a block on frictionless ice. Starting from rest, the block travels a distance din

sweet [91]

Answer:

<em>The distance is now 4d</em>

Explanation:

<u>Mechanical Force</u>

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = m.a

Where a is the acceleration of the object.

The acceleration can be calculated by solving for a:

$\displaystyle a=\frac{F}{m}$

Once we know the acceleration, we can calculate the distance traveled by the block as follows:

$\displaystyle d = vo.t+\frac{at^2}{2}$

If the block starts from rest, vo=0:

$\displaystyle d = \frac{at^2}{2}$

Substituting the value of the acceleration:

$\displaystyle d = \frac{\frac{F}{m}t^2}{2}$

Simplifying:

$\displaystyle d = \frac{Ft^2}{2m}$

When a force F'=4F is applied and assuming the mass is the same, the new acceleration is:

$\displaystyle a'=\frac{4F}{m}$

And the distance is now:

$\displaystyle d' = \frac{4Ft^2}{2m}$

Dividing d'/d:

$\displaystyle \frac{d' }{d}=\frac{\frac{4Ft^2}{2m}}{\frac{Ft^2}{2m}}$

Simplifying:

$\displaystyle \frac{d' }{d}=4$

Thus:

d' = 4d

The distance is now 4d

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2 years ago

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Fantom [35]

Marble i think not quite sure doe

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defon

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2 years ago

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