If you were to mix 10 grams of salt into 100 grams of water, how much will the total mass of the solution be?

Holding onto a tow rope moving parallel to a frictionless ski slope, a 70.1 kg skier is pulled up the slope, which is at an angl

lara [203]

Answer:

given,

mass of the skier = 70.1 Kg

angle with horizontal, θ = 8.6°

magnitude of the force,F = ?

a) Applying newton's second law

$m g sin\theta - F_{rope} = ma$

velocity is constant, a = 0

$m g sin\theta - F_{rope} =0$

$F_{rope} = m g sin\theta$

$F_{rope} = 70.1\times 9.8\times sin 8.6^0$

$F_{rope}= 102.73\ N$

b) now, when acceleration, a = 0.135 m/s²

$F_{rope}-m g sin\theta = ma$

velocity is constant, a = 0.135 m/s₂

$F_{rope} = m g sin\theta+ma$

$F_{rope} = 70.1\times 9.8\times sin 8.6^0+70.1\times 0.135$

$F_{rope}= 112.19\ N$

7 0

3 years ago

4. A student measures a temperature several times. The readings lie between 29.6 and 30.2 K. This

Alona [7]

Answer:

(C) apparently written incorrectly - it should be 29.9 +- .3 K

(read 29.9 plus or minus .3 K)

8 0

3 years ago

write a one or two summary paragraph discussing this experiment and the results use tye following questions 1 according to your

Monica [59]

Answer:heed

Explanation:

heed knows how to divide.

8 0

3 years ago

A horizontal force of 400 N is exerted on a 2.0-kg ball as it rotates (at

frutty [35]

Answer:

the speed of the ball is 10 m/s

Explanation:

Given;

magnitude of exerted force, F = 400 N

mass of the ball, m = 2 kg

radius of the circle, r = 0.5

The speed of the ball is calculated by applying centripetal force formula;

$F = \frac{mv^2}{r} \\\\v^2 = \frac{Fr}{m}\\\\v = \sqrt{\frac{Fr}{m}}\\\\ v = \sqrt{\frac{400*0.5}{2}}\\\\v = 10 \ m/s$

Therefore, the speed of the ball is 10 m/s

6 0

3 years ago

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If the coefficient of static friction is 0.357, and the same ladder makes a 58.0° angle with respect to the horizontal, how far

zavuch27 [327]

Answer: d= 0.57* l

Explanation:

We need to check that before ladder slips the length of ladder the painter can climb.

So we need to satisfy the equilibrium conditions.

So for ∑Fx=0, ∑Fy=0 and ∑M=0

We have,

At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal

At the top of ladder, N₂ acting horizontal

And Between somewhere we have the weight of painter acting downward equal to= mg

So, we have N₁=mg

and also mg*d*cosФ= N₂*l*sin∅

So,

d= $\frac{N2}{mg}*l$ * tan∅

Also, we have f₁=N₂

As f₁= чN₁

So f₁= 0.357 * 69.1 * 9.8

f₁= 241.75

Putting in d equation, we have

d= $\frac{241.75}{69.1*9.8} *l$ * tan 58

d= 0.57* l

So painter can be along the 57% of length before the ladder begins to slip

3 0

3 years ago

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