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SVEN [57.7K]
2 years ago
10

Help me pls, I need it now​

Physics
1 answer:
posledela2 years ago
3 0

6.) a

7.) d

8.) b

9.) b

10.) a

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6. What is the molar mass of H2O?
kotykmax [81]

Answer:

a) 18.015

Explanation:

I Googled it, hope this helped

7 0
3 years ago
A 1.1-kg object is suspended from a vertical spring whose spring constant is 120 N/m. (a) Find the amount by which the spring is
andriy [413]

Answer:

e = 0.0898m

v = 2.07m/s

Explanation:

a) According to Hooke's law

F = ke

e is the extension

k is the spring constant

Since F = mg

mg = ke

e = mg/k

Substitute the given value

e = 1.1(9.8)/120

e = 10.78/120

e = 0.0898m

Hence it is stretched by 0.0898m from its unstrained length

2) Total Energy = PE+KE+Elastic potential

Total Energy = mgh +1/2mv²+1/2ke²

Substitute the given value

5.0= 1.1(9.8)(0.2)+1/2(1.1)v²+1/2(120)(0.0898)²

Solve for v

5.0 = 2.156+0.55v²+0.48338

5.0-2.156-0.48338= 0.55v²

2.36 =0.55v²

v² = 2.36/0.55

v² = 4.29

v ,= √4.29

v = 2.07m/s

Hence the required velocity is 9.28m/s

4 0
3 years ago
A launched hopper reach to 1.20 m maximum height. How much is it’s launch velocity?
garri49 [273]

The launch velocity is 4.8 m/s

Explanation:

We can solve this problem by applying the law of conservation of energy. In fact, the mechanical energy of the hopper (equal to the sum of the potential energy + the kinetic energy) is conserved. So we can write:

U_i +K_i = U_f + K_f

where:

U_i is the initial potential energy, at the bottom

K_i is the initial kinetic energy, at the bottom

U_f is the final potential energy, at the top

K_f is the final kinetic energy, at the top

We can rewrite the equation as:

mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2

where:

m is the mass of the hopper

g=9.8 m/s^2 is the acceleration of gravity

h_i = 0 is the initial height

u is the launch speed of the hopper

h_f = 1.20 m is the maximum altitude reached by the hopper

v = 0 is the final speed (which is zero when the hopper reaches the maximum height)

Solving the equation for u, we find the launch speed of the hopper:

u=\sqrt{2gh_g}=\sqrt{2(9.8)(1.20)}=4.8 m/s

Learn more about kinetic energy and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647  

brainly.com/question/10770261  

#LearnwithBrainly

4 0
3 years ago
In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole.
Fiesta28 [93]

Answer:

6.0 m/s

Explanation:

According to the law of conservation of energy, the total mechanical energy (potential, PE, + kinetic, KE) of the athlete must be conserved.

Therefore, we can write:

KE_i+PE_i =KE_f+PE_f

or

\frac{1}{2}mu^2+0=\frac{1}{2}mv^2+mgh

where:

m is the mass of the athlete

u is the initial speed of the athlete (at the bottom)

0 is the initial potential energy of the athlete (at the bottom)

v = 0.80 m/s is the final speed of the athlete (at the top)

g=9.8 m/s^2 is the acceleration due to gravity

h = 1.80 m is the final height of the athlete (at the top)

Solving the equation for u, we find the initial speed at which the athlete must jump:

u=\sqrt{v^2+2gh}=\sqrt{0.80^2+2(9.8)(1.80)}=6.0 m/s

4 0
3 years ago
Who live in a pineapple under the sea?
saw5 [17]

Answer:

SpongeBob SquarePants

5 0
3 years ago
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