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MAXImum [283]
3 years ago
7

A velocity-time graph is shown below: (2 points)

Physics
1 answer:
BabaBlast [244]3 years ago
5 0

Answer:

2m/s²

Explanation:

  • V=Vstart+at
  • rewrite that to find a so a=(V-Vstart)/t

part A of graph

  • a=(20m/s–0m/s)/5s
  • a=4m/s²

part B of graph

  • a=(0m/s–0m/s)/5s
  • a=0m/s²

the average between the two is both answers added divided by the number of answers

  • (4m/s²–0m/s²)/2
  • 4m/s²/2
  • 2m/s²
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A book weighing 15 N sits on a desk. What is the force the desk is exerting back on the book?
bazaltina [42]

Answer:

the book is exerting a force of 15N on the table as we know that every action has an equal but opposite reaction so there must be a force acting on it in the opposite direction the other force normal force or we can say it perpendicular force that is defined as a contact force perpendicular to the contact surface that prevent two objects from passing over one another

Explanation:

i hope it will help you

3 0
4 years ago
When we experience positive "g forces", it is as if we have become...
zhenek [66]

Answer:

heavier

Explanation:

7 0
3 years ago
Someone help with this pls
Elan Coil [88]
The answer would be 91.44
3 0
3 years ago
Read 2 more answers
Dos cargas de 1 pC estan a 1mm, la fuerza de interaccion entre dichas cargas es?​
Naddik [55]

Answer:

La fuerza de interacción entre dichas partículas es 8.988\times 10^{-9}\,N.

Explanation:

Asumamos que las dos cargas son puntuales, puesto que la fuerza de interacción es netamente electrostática, es determinada por la Ley de Coulomb:

F = \frac{k\cdot q_{1}\cdot q_{2}}{r^{2}} (1)

Donde:

k - Constante electrostática, medida en newton-metros cuadrados por Coulomb cuadrado.

q_{1}, q_{2} - Carga eléctrica, medida en Coulomb.

r - Distancia entre las partículas, medida en metros.

Si sabemos que k = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}, q_{1} = q_{2} = 1\times 10^{-12}\,C y r = 1\times 10^{-3}\,m, entonces la fuerza de interacción entre ambas partículas es:

F = \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (1\times 10^{-12}\,C)^{2}}{(1\times 10^{-3}\,m)^{2}}

F = 8.988\times 10^{-9}\,N

La fuerza de interacción entre dichas partículas es 8.988\times 10^{-9}\,N.

6 0
3 years ago
A karate expert breaks a stack of bricks with his bare hand. If the force applied is 520 newtons and the impact time is 5.0 × 10
Scrat [10]
This question could be solved with this formula:
F . ΔT
From which F= The amount of force applied
and ΔT = the impact time ins seconds

So, the Value of impulse would be:
520 N x 5. x 10^-4 seconds

= 0.052 N x 5

= 0.26 Newton Seconds
6 0
4 years ago
Read 2 more answers
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