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GaryK [48]
2 years ago
6

Believing that Nick will not be a good waiter because he is 60 years old is an example of O Stereotype O Prejudice Situational a

ttribution All of the above​
Physics
1 answer:
Helen [10]2 years ago
4 0
Ya so quid fuh doil em yous
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Describe two ways unbalanced forces help you in your day to<br> day life.
Alexxandr [17]

Answer:

  1. we need unbalance force to lift objects
  2. we need unbalance force to drag objects
6 0
3 years ago
True or false the hotter the star the higher is absolute magnitude?
liubo4ka [24]

Answer:

true

Explanation:

6 0
3 years ago
A mass on a horizontal surface is connected to the spring and pulled to the right along the surface stretching the spring by 25
solniwko [45]

Answer:

320 N/m

Explanation:

From Hooke's law, we deduce that

F=kx where F is applied force, k is spring constant and x is extension or compression of spring

Making k the subject of formula then

k=\frac {F}{x}

Conversion

1m equals to 100cm

Xm equals 25 cm

25/100=0.25 m

Substituting 80 N for F and 0.25m for x then

k=\frac {80}{0.25}=320N/m

Therefore, the spring constant is equal to 320 N/m

3 0
3 years ago
What happens when the falling object reaches its terminal velocity? It will stop _________
lora16 [44]

Answer:

LOL i don't even know

Explanation:

7 0
2 years ago
A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o
iVinArrow [24]
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: v=15 m/s
- radius of the hill: r=100 m

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car W=mg (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, m \frac{v^2}{r}, so we can write:
mg-N=m \frac{v^2}{r} (1)
By rearranging the equation and substituting the numbers, we find N:
N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
mg=m \frac{v^2}{r}
from which we find
v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s
7 0
3 years ago
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