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Temka [501]
3 years ago
5

PLEASE ANSWER I DONT HAVE THAT MUCH TIME

Physics
1 answer:
ollegr [7]3 years ago
7 0

Answer:

its paper

Explanation:

hope it helps

ace ur test !

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A ball is thrown straight up with a speed of 30
konstantin123 [22]

Answer:

A ball is thrown straight up with a speed of 30

m/s. What is the maximum height reached by

the ball?

4 0
3 years ago
A 1200 Kg car rounds a corner of radius r = 45m. If the coefficient of static friction between the ties and the road is us = 0.8
Vaselesa [24]

Answer:

The greatest speed of the car is 19.36m/s

Explanation:

The maximum speed the car will attain without skidding is given by:

F= uN = umg ...eq1

But F = mv^2/r

mv^2/r = umg

Dividing both sides by m, leaves you with:

V= Sqrt(ugr)

Where u = coefficient of static friction

g = acceleration due to gravity

r = raduis

Given:

U = 0.82

r=0.82

g= 9.8m/s

V = Sqrt(0.82 × 9.8 × 45)

V = Sqrt(374.85)

V = 19.36m/s

5 0
4 years ago
Ted Clubber Lang. A hook in boxing primarily involves horizontal flexion of the shoulder while maintaining a constant angle at t
il63 [147K]

Answer:

15 m/s or 1500 cm/s

Explanation:

Given that

Speed of the shoulder, v(h) = 75 cm/s = 0.75 m/s

Distance moved during the hook, d(h) = 5 cm = 0.05 m

Distance moved by the fist, d(f) = 100 cm = 1 m

Average speed of the fist during the hook, v(f) = ? cm/s = m/s

This can be solved by a very simple relation.

d(f) / d(h) = v(f) / v(h)

v(f) = [d(f) * v(h)] / d(h)

v(f) = (1 * 0.75) / 0.05

v(f) = 0.75 / 0.05

v(f) = 15 m/s

Therefore, the average speed of the fist during the hook is 15 m/s or 1500 cm/s

6 0
3 years ago
Explain three characteristic of a uniform electric field​
Trava [24]

Answer:

• They depend solely on the load that generates it

• Two or more electrical charges interact, which can be positive or negative

• The energy source is based on the electrical voltage

3 0
3 years ago
In still​ water, a boat averages 18 18 miles per hour. it takes the same amount of time to travel 16 miles 16 miles ​downstream,
Vladimir79 [104]
<span>The current is 6 miles per hour.
   Let's create a few equations:
 Traveling with the current:
 (18 + c)*t = 16

   Traveling against the current:
 (18 - c)*t = 8

   Let's multiply the 2nd equation by 2
 (18 - c)*t*2 = 16

   Now subtract the 1st equation from the equation we just doubled.
 (18 - c)*t*2 = 16
 (18 + c)*t = 16

   (18 - c)*t*2 - (18 + c)*t = 0
 Divide both sides by t
 (18 - c)*2 - (18 + c) = 0

   Now solve for c
 (18 - c)*2 - (18 + c) = 0
 36 - 2c - 18 - c = 0
 36 - 2c - 18 - c = 0
 18 - 3c = 0
 18 = 3c
  6 = c

   So the current is 6 mph.
   Let's verify that.
 (18 + 6)*t = 16
 24*t = 16
 t = 16/24 = 2/3

   (18 - 6)*t = 8
 12*t = 8
 t = 8/12 = 2/3

   And it's verified.</span>
4 0
3 years ago
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