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tigry1 [53]
4 years ago
7

Do you think it would be possible to get your maximum heart rate ? Why or why not

Physics
1 answer:
Mazyrski [523]4 years ago
8 0

Answer:

Speaking hypothitically, since I dont have a lot of knowledge around this subject, but yes i think it's possible, and probably not as rare as one might think.

Explanation:

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A toy car is given an initial velocity of 5.0 m/s and experiences a constant acceleration of 2.0 m/s656-03-02-00-00_files/i02900
Whitepunk [10]
It would be 17 m/s

If we use

V2 = V1 + a*t
Sub in 5 for v1
2m/s*2 for a
And
6 for t
That should give you the answer.
5 0
3 years ago
Read 2 more answers
10 points!! If you help!!!
marta [7]

For Mass

K.E = (1/2*mv^2)

Explanation:

Kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s2.

4 0
3 years ago
If the net external force acting on a system is zero , then the total momentum of the system is zero
Neporo4naja [7]
If net external force acting on the system is zero, momentum is conserved. That means, initial and final momentum are same → total momentum of the system is zero.
8 0
4 years ago
Read 2 more answers
A baseball player hits a baseball with a bat. The mass of the ball is 0.05 kilograms. The ball accelerates at 2,000 meters per s
AlexFokin [52]

Answer:

550

Explanation:

8 0
3 years ago
A hockey puck moving at 0.4600 m/s collides with another puck that was at rest. The pucks have equal mass. The first puck is def
Sladkaya [172]

Answer:

Speed =0.283m/ s

Direction = 47.86°

Explanation:

Since it is a two dimensional momentum question with pucks having the same mass, we derive the momentum in xy plane

MU1 =MU2cos38 + MV2cos y ...x plane

0 = MU2sin38 - MV2sin y .....y plane

Where M= mass of puck, U1 = initial velocity of puck 1=0.46, U2 = final velocity of puck 1 =0.34, V2 = final velocity of puck 2, y= angular direction of puck2

Substitute into equation above

.46 = .34cos38 + V2cos y ...equ1

.34sin38 = V2sin y...equ2

.19=V2cos Y...x

.21=V2sin Y ...y

From x

V2 =0.19/cost

Sub V2 into y

0.21 = 0.19(Sin y/cos y)

1.1052 = tan y

y = 47.86°

Sub Y in to x plane equ

.19 = V2 cos 47.86°

V2=0.283m/s

7 0
4 years ago
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