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Gre4nikov [31]
3 years ago
5

A spring has a spring constant of 100 N/m , and the mass hanging on it is 0.50 kg. What is the period of the spring's motion?

Physics
1 answer:
Basile [38]3 years ago
8 0

Answer:

0.44 sec

Explanation:

k = 100 N/m, m = 0.50 kg

Period

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You took a running leap off a high diving platform. You were running at 3.1 m/s and hit the water 2.4 seconds later. How high wa
rusak2 [61]

The distance you free-fall from rest is  D = (1/2) (g) (T²) <== memorize this

Height of the platform = (1/2) (9.8 m/s²) (2.4 sec)²

Height = (4.9 m/s²) (5.76 s²)

Height = (4.9/5.76) meters

Height = 28.2 meters (a VERY high platform ... about 93 ft off the water !)

Without air-resistance, your horizontal speed doesn't change.  It's constant.  Traveling 3.1 m/s for 2.4 sec, you cover (3.1 m/s x 2/4 s) = 7.4 m horizontally.

7 0
4 years ago
An object on Earth weighs 150 N. What is its mass?
Alex777 [14]

Answer:

\tt \: mass \:  =  \frac{weight}{acceleration \: due \: to \: gravity}

\longrightarrow \tt \:  \frac{150}{10}

\longrightarrow \boxed{ \tt{15 \: kg}}

  • Our final answer is 15 kg .

----------- HappY LearninG<3 ------------

3 0
2 years ago
A ranger in a national park is driving at 56 km/h when a deer jumps onto the road 65 m ahead of the vehicle. After a reaction ti
vagabundo [1.1K]

Answer:

 t = 1.58 s

Explanation:

given,

Speed of ranger, v = 56 km/h

                            v = 56 x 0.278 = 15.57 m/s

distance, d = 65 m

deceleration,a = 3 m/s²

reaction time = ?

using stopping distance formula

d = v. t + \dfrac{v^2}{2a}

t = \dfrac{d}{v} -\dfrac{v}{2a}

t is the reaction time

t = \dfrac{65}{15.57} -\dfrac{15.57}{2\times 3}

 t = 1.58 s

hence, the reaction time of the ranger is equal to 1.58 s.

3 0
3 years ago
PLEASE HELPPPP!!!<br> Density is the amount of mass per unit volume.<br><br><br> True or False
Alexxandr [17]

Answer:

true :D

Explanation:

4 0
3 years ago
A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light supporting cable
kompoz [17]

Answer:

4.44s

Explanation:

A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light supporting cables for the swing is 4.9 m, how long does it take for each complete back-and-forth swing? Assume that the child and swing set are very small compared to the length of the cables

since the mass of the child and that of the swing is negligible, the masses wont be involved in the calculation

T=2π√L/g

g=acceleration due to gravity which is 9.81m/s2

the length of the supporting cable is 4.9m

T the period

period is the time required to make a complete oscillation

T=2*π√4.9/9.81

T=2*π*0.706

T=4.44s

4.44s

5 0
4 years ago
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