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Drupady [299]
3 years ago
14

Kramer goes bowling and decides to employ the force of gravity to "pick up a spare." He rolls the 7.0-kg bowling ball very slowl

y so that it comes to rest a center-to-center distance of 0.10 m from the one remaining 1.5-kg bowling pin. 1) Determine the force of gravity between the ball and the pin and comment on the efficacy of the technique. Treat the ball and pin as point objects for this problem. (Express your answer to two significant figures.) nN
Physics
1 answer:
Rzqust [24]3 years ago
3 0

Answer:

70.035nN

Explanation:

M₁ = 7.0kg

m₂ = 1.5kg

r = 0.1m

G = 6.67*10⁻¹¹ Nm²/kg

Gravitational force (F) of an object is directly proportional to the product masses and inversely proportional to the square of their distance apart.

F = GM₁M₂ / r²

F = (6.67*10^-11 * 7.0 * 1.5) / (0.1)²

F = 7.0035*10^-10 / 0.01

F = 7.0035 * 10^-8 N

F = 70.035nN

The gravitational force acting on the bowling ball is 70.035nN.

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Find a unit vector normal to the plane containing Bold u equals 2 Bold i minus Bold j minus 3 Bold k and Bold v equals negative
Nastasia [14]

Answer:

\hat{w}=\dfrac{\vec{5i+13j-k}}{13.92}

Explanation:

It is given that,

\vec{u}=2i-j-3k

\vec{v}=-3i+j-2k

Taking the cross product of v and v such that,

\vec{w}=u\times v

\vec{w}=(2i-j-3k)\times (-3i+j-2k)

\vec{w}=5i+13j-k

|w|=\sqrt{5^2+13^2(-1)^2}

|w| = 13.92

Let \hat{w} is the unit vector normal to the plane containing u and v. So,

\hat{w}=\dfrac{\vec{w}}{|w|}

\hat{w}=\dfrac{\vec{5i+13j-k}}{13.92}

Hence, this is the required solution.

7 0
3 years ago
Do y’all know the answers to them all the choices are at the drop down box.!!!
Alborosie

the second option is your answer to what a liquid is

4 0
3 years ago
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You walk 100m due north. You then turn and walk 55m due east. You then make another turn and walk 12m due south. What is the res
lord [1]

Answer:

Explanation:

Important here is to know that due north is a 90 degree angle, due east is a 0 degree angle, and due south is a 270 degree angle. Then we find the x and y components of each part of this journey using the sin and cos of the angles multiplied by each magnitude:

A_x=100cos90\\A_x=0\\B_x=55cos0\\B_x=55\\C_x=12cos270\\C_x=55

Add them all together to get the x component of the resultant vector, V:

V_x=55

Do the same to find the y components of the part of this journey:

A_y=100sin90\\A_y=100\\B_y=55sin0\\B_y=0\\C_y=12sin270\\C_y=-12

Add them together to get the y component of the resultant vector, V:

V_y=88

One thing of import to note is that both of these components are positive, so the resultant angle lies in QI.

We find the final magnitude:

V_{mag}=\sqrt{55^2+88^2} and, rounding to 2 sig dig's as needed:

V_{mag}= 1.0 × 10² m; now for the direction:

\theta=tan^{-1}(\frac{88}{55})= 58°

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2 years ago
When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees
dolphi86 [110]

Answer:

a) 1.725*10^5 N

b) 3.83*10^3 N

c) i) 173.24 kN

c) ii) 4.57 kN

Explanation:

See the attachment for calculations

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2 years ago
A car was moving a 14 m/s. After 30 seconds, it’s speed increased to 20 m/s. What was its acceleration during this time?
Anuta_ua [19.1K]

Answer:

acceleration = 0.2 m/s/s

Explanation:

initial velocity u = 14

final velocity v = 20

time = 30

acceleration = ?

v = u + at

20 = 14 + 30a

30a = 6

a = 0.2 m/s/s

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2 years ago
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