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rodikova [14]
3 years ago
12

Kinetic energy problemset

Physics
1 answer:
Sonja [21]3 years ago
8 0

Answer:

KE = 4 mv2 m = 2xKE valami. V m.

Explanation:

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If a ball is thrown vertically upward with a velocity of 128 ft/s, then its height after t seconds is s = 128t − 16t2.
Fed [463]

Answer:

a) 256 ft

b) 32 ft/s

c) -32 ft/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32 ft/s²

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128^2}{2\times -32}\\\Rightarrow s=256\ ft

Maximum height = 256 ft

When s = 240 ft

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -32\times 240+128^2}\\\Rightarrow v=32\ ft/s

Speed of the ball at 240 ft is 32 ft/s while going up

Now this will be the velocity of the ball when it reaches 240 ft, which will be considered as the initial velocity

v=u+at\\\Rightarrow 0=32-32t\\\Rightarrow t=\frac{-32}{-32}=1\ s

Now, initial velocity will be considered as zero

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 32\times 16+0^2}\\\Rightarrow v=32\ ft/s

Speed of the ball at 240 ft is -32 ft/s while going down

3 0
3 years ago
A crew member is walking on a tugboat that is pulling a barge. The tugboat is moving at a constant speed upstream in a river tha
Marianna [84]

Answer:

v₁

Explanation:

A crew member is walking on a tugboat that is pulling a barge. The tugboat is moving at a constant speed upstream in a river that has a constant downstream current. The velocity of the crew member with respect to the tugboat is v₁. What is the velocity of the crew member with respect to the barge?

velocity is the change in displacement per time.

the velocity of the crew member with respect to the barge is v₁ since the barge is moving in the velocity of tugboat

4 0
3 years ago
Hi please help me with this question. ..
natta225 [31]

Explanation:

For projectile motion, use constant acceleration equation:

Δx = v₀ t + ½ at²

where Δx is the displacement,

v₀ is the initial velocity,

a is the acceleration,

and t is time.

Both objects are projected upward with velocity u.  The second object is thrown after a time t₀.

For the first object:

Δx = u t + ½ (-g) t²

Δx = u t − ½g t²

For the second object:

Δx = u (t−t₀) + ½ (-g) (t−t₀)²

Δx = u (t−t₀) − ½g (t−t₀)²

Assuming the objects meet, the displacements will be equal:

u t − ½g t² = u (t−t₀) − ½g (t−t₀)²

u t − ½g t² = u (t−t₀) − ½g (t² − 2tt₀ + t₀²)

u t − ½g t² = u t − u t₀ − ½g t² + g tt₀ − ½g t₀²

0 = -u t₀ + g tt₀ − ½g t₀²

0 = -u + g t − ½g t₀

g t = u + ½g t₀

t = u/g + t₀/2

8 0
4 years ago
Osteoporosis is a condition that makes bones more
erastova [34]

the answer for this question is c

7 0
3 years ago
a car has a velocity of 10ms-1. it accelerates at 0.2 ms-2 for half minute find the total distance travelled and final velocity
Readme [11.4K]

Answer:

Final velocity = 16 m/s

Total distane travelled = 390 m

Explanation:

We can use equation of motion to solve this:

v = u + at \\ v = 10 + 0.2(30) \\ v = 16 {ms}^{ - 1}

s = ut +  \frac{1}{2} a {t}^{2}  \\ s = 10(30) +  \frac{1}{2} (0.2) {(30)}^{2}  \\ s = 390m

5 0
3 years ago
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