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andrezito [222]

3 years ago

15

A 100 N box is placed on an inclined plane that makes a 30° angle with the horizontal. Calculate the component of gravitational

force that acts down the inclined plane.
A. 25 N
B. 50 N
C. 87 N
D. 100 N

1 answer:

nika2105 [10]3 years ago

6 0

B is the Answer 50 N

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In a fast-pitch softball game the pitcher is impressive to watch, as she delivers a pitch by rapidly whirling her arm around so

svetlana [45]

Answer:

(a). The magnitude of the total acceleration of the ball is 239.97 m/s².

(b). The angle of the total acceleration relative to the radial direction is 11.0°

Explanation:

Given that,

Radius of the circle = 0.681 m

Angular acceleration = 67.7 rad/s²

Angular speed =18.6 rad/s

We need to calculate the centripetal acceleration of the ball

Using formula of centripetal acceleration

$a_{c}=\omega^2\times r$

Put the value into the formula

$a_{c}=(18.6)^2\times0.681$

$a_{c}=235.5\ m/s^2$

We need to calculate the tangential acceleration of the ball

Using formula of tangential acceleration

$a_{t}=r\alpha$

Put the value into the formula

$a_{t}=0.681\times67.7$

$a_{t}=46.104\ m/s^2$

(a). We need to calculate the magnitude of the total acceleration of the ball

Using formula of total acceleration

$a=\sqrt{a_{c}^2+a_{t}^2}$

Put the value into the formula

$a=\sqrt{(235.5)^2+(46.104)^2}$

$a=239.97\ m/s^2$

(b). We need to calculate the angle of the total acceleration relative to the radial direction

Using formula of the direction

$\theat=\tan^{-1}(\dfrac{a_{t}}{a_{c}})$

Put the value into the formula

$\theta=\tan^{-1}(\dfrac{46.104}{235.5})$

$\theta=11.0^{\circ}$

Hence, (a). The magnitude of the total acceleration of the ball is 239.97 m/s².

(b). The angle of the total acceleration relative to the radial direction is 11.0°

5 0

3 years ago

A 2.5 kg table can hold kg of weight. what is its structural efficiency?

Lorico [155]

It's (whatever number comes after 'hold') divided by 2.5 .

6 0

2 years ago

A 4.4 kg marble (really big heavy marble) is accelerating down an incline. When it reaches level ground it slows down to a stop

KengaRu [80]

#A

Mass=4.4kg

acceleration=-1.74m/s^2

Use newtons second law

$\\ \rm\longmapsto Force=ma$

$\\ \rm\longmapsto Force=4.4(-1.74)$

$\\ \rm\longmapsto Force=-7.656N$

#B

initial velocity=u

Final velocity=v=0

Acceleration=a=-1.74m/s^2

Time=t=1.27s

$\\ \rm\longmapsto a=\dfrac{v-u}{t}$

$\\ \rm\longmapsto u=v-at$

$\\ \rm\longmapsto u=0-(-1.74)(1.27)$

$\\ \rm\longmapsto u=1.74(1.27)$

$\\ \rm\longmapsto u=2.2m/s$

4 0

2 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. Simu

salantis [7]

Answer: pick A moves 10.2 metre

Explanation: since they were given a quick push simultaneously the product of their distance moved and velocity would be equal.

Let the distance moved by puck B x then that of puck A would be 18-x.

So therefore we have that,

1.3*(18-x) = 1.7*x

23.4 - 1.3x = 1.7x

23.4=3x

Making x subject of formula we have

x = 23.4/3 = 7.8m

But the distance moved by puck A is 18-x. So therefore,

18-7.8=10.2meter

6 0

2 years ago

Read 2 more answers

I NEED HELP ASAP

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8 0

2 years ago