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4 years ago

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A gardener pushes her 90 kg wheelbarrow in a straight line out of her garage and across the yard in 18 s. If her average velocit

y is 1.5 m/s, what is her displacement?
A. 0.8 km
B. 12 km toward the garden
C. 18 m toward the garage
D. 27 m away from the garage

1 answer:

Hunter-Best [27]4 years ago

3 0

The answer is D, because displacement equals average velocity times time. 1.5 m/s times 18s equals 27 meters.

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user100 [1]

Answer:

a = - 50 [m/s²]

Explanation:

To solve this problem we simply have to replace the values supplied in the given equation.

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Now replacing we have:

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3 years ago

See the diagram above. Which of the following is the best prediction of what would happen if you increased the distance in secti

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By the definition of wavelength, the answer is the letter D, the wavelength would decrease.

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Now, if we increase the distance of the crests, by the definition shown above, we will increase the wavelength.

Therefore, the answer is letter D, the wavelength would increase.

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3 years ago

: A 70 kg man and a 12 kg sled are on the frictionless ice of a frozen lake, 25 m apart but connected by a rope of negligible ma

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Answer:

$x_1 = 3.74m$

Explanation:

given,

mass of man = 70 kg

mass of sled = 12 kg

F = m a_s

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$a_s = 0.68\ m/s^2$

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$a_m = \dfrac{8.2}{70}$

$a_m = 0.12\ m/s^2$

$x_1+x_2 = 25$

$\dfrac{1}{2}a_ct^2+ \dfrac{1}{2}a_mt^2 = 25$

$(a_c+a_m)t^2=50$

$(0.12+0.68)t^2=50$

$t = \sqrt{\dfrac{50}{0.8}}$

t = 7.90 s

$x_1 = \dfrac{1}{2}a_ct^2$

$x_1 = 0.5\times 0.12 \times 7.90^2$

$x_1 = 3.74m$

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3 years ago

The diagrams show objects’ gravitational pull toward each other. Which statement describes the relationship between diagram X an

creativ13 [48]

' C ' is the only correct statement on the list. We don't know anything about diagram-x or diagram-y because we can't see them.

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3 years ago

Read 2 more answers

A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.0 s after being hit. Then 2.5 s afte

vaieri [72.5K]

Answer:

Part a)

$H = 44.1 m$

Part b)

$y = 13.48 m$

Part c)

$d = 8.86 m$

Explanation:

Part a)

As we know that ball will reach at maximum height at

t = 3 s

now we will have

$t = \frac{v sin\theta}{g}$

now we have

$3 = \frac{vsin\theta}{9.8}$

$v sin\theta = 29.4 m/s$

Now maximum height above ground is given as

$H = \frac{v^2sin^2\theta}{2g}$

$H = \frac{29.4^2}{2(9.8)}$

$H = 44.1 m$

Part b)

Height of the fence is given as

$y = (vsin\theta) t - \frac{1}{2}gt^2$

$y = (29.4)(5.5) - \frac{1}{2}(9.8)(5.5^2)$

$y = 13.48 m$

Part c)

As we know that its horizontal distance moved by the ball in 5.5 s is given as

$x = v_x t$

$97.5 = v_x (5.5)$

$v_x = 17.72 m/s$

now total time of flight is given as

$T = 3 + 3 = 6 s$

so range is given as

$R = v_x T$

$R = (17.72)(6)$

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so the distance from the fence is given as

$d = 106.4 - 97.5$

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4 years ago