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nalin [4]
3 years ago
10

When jogging outside you accidentaly bumb into a curb. Your feet stop but your body continues to movr foward and you end up on t

he ground. Which law of motion is being described in this scenario?
A.Law of universal gravitation
B.Newton's first law of motion
C.Law of conservation of energy
D. Newton's second law of Motion
Physics
2 answers:
sattari [20]3 years ago
8 0
B.) Newtons first law of motion
Monica [59]3 years ago
7 0
It is his first law of motion it states "Every object persists in its state of rest or uniform motion in a straight line unless it is compelled to change that state by forces impressed on it"<span />
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Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after
s2008m [1.1K]

Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after the objects collide and bounce apart?

a-velocity

b-mass

c-momentum

d-direction

Answer:

b. Mass

Explanation:

This question has to do with the principle of the law of conservation of momentum which states that the momentum of a system remains constant if no external force is acting on it.

As the question states, two objects collide with each other and eventually bounce apart, so their momentum may not be conserved but the mass of the objects is constant for each non-relativistic motion. Because of this, the mass of each object prior to the collision would be the same as the mass after the collision.

Therefore, the correct answer is B. Mass.

6 0
4 years ago
Read 2 more answers
(a) The stick is supported by a sharp point at the middle. On the left side, a weight of 100 g is suspended at 40 cm from the mi
Jet001 [13]

Answer:

20cm

Explanation:

Hello!

remember that the condition for a body to be at rest is that the sum of its moments and its forces be zero,

To solve this problem you must draw the free body diagram of the stick (attached image) and sum up moments at point 0 (where the sharp is located), which results in the following equation

(100g)(40cm)=x(200g)

X=\frac{(100)(40)}{(200)} =20cm

6 0
4 years ago
A 0.1 kg toy contains a compressed spring. when the spring is released the toy fly 0.45 m upwards from ground level before falli
trapecia [35]

The speed of the toy when it hits the ground is 2.97 m/s.

The given parameters;

  • mass of the toy, m = 0.1 kg
  • the maximum height reached by the, h = 0.45 m

The speed of the toy before it hits the ground will be maximum. Apply the principle of conservation of mechanical energy to determine the maximum speed of the toy.

P.E = K.E

mgh_{max} = \frac{1}{2} mv_{max}^2\\\\gh_{max} = \frac{1}{2} v_{max}^2\\\\v_{max}^2= 2gh_{max}\\\\v_{max} = \sqrt{2gh_{max}}

Substitute the given values and solve the speed;

v_{max} = \sqrt{2\times 9.8 \times 0.45} \\\\v_{max} = 2.97 \ m/s

Thus, the speed of the toy when it hits the ground is 2.97 m/s.

Learn more here: brainly.com/question/7562874

7 0
3 years ago
If a ball leaves the ground with a vertical velocity of 5.46 m/s, how long does it take
kifflom [539]

Answer:

0.557 s

Explanation:

Given:

v₀ = 5.46 m/s

v = 0 m/s

a = -9.8 m/s²

Find: t

v = at + v₀

0 m/s = (-9.8 m/s²) t + 5.46 m/s

t = 0.557 s

8 0
3 years ago
A ball of mass m is thrown into the air in a 45° direction of the horizon, after 3 seconds the ball is seen in a direction 30° f
Rzqust [24]

Answer:

Velocity (magnitude) is 98.37 m/s

Explanation:

We use the vertical component of the initial velocity, which is:

v_{0y}=v_0*sin(45)=\frac{\sqrt{2} }{2}v_0

Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):

v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t

Now we need to find v_y as a function of v_0. We use the horizontal velocity, which is always the same as follow:

v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\

We know the angle at 3 seconds:

v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}

Substitute  v_{t=3} in  v_x and then solve for  v_y

\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0

With this expression we go back to the kinematic equation and solve it for initial speed

\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s

3 0
4 years ago
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