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stepladder [879]
3 years ago
11

A control clinic offers a program that guarantees a weight loss of up to 0.46 kg in one week. Express the weight loss in a ratio

of milligrams per second.
Physics
1 answer:
Mice21 [21]3 years ago
4 0

Answer:

0.76 mg/s

Explanation:

0.46 kg/week × (1 week / 7 days) × (1 day / 24 hrs) × (1 hr / 3600 s) × (1000 g/kg) × (1000 mg/g) = 0.76 mg/s

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A government agency estimated that air bags have saved over 14,000 lives as of April 2004 in the United States. (They also state
balu736 [363]

To solve this problem it is necessary to apply the concepts related to momentum, momentum and Force. Mathematically the Impulse can be described as

I = F*t

Where,

F= Force

t= time

At the same time the moment can be described as a function of mass and velocity, that is

P = m\Delta v \rightarrow P=m(v_1-v_2)

Where,

m = mass

v = Velocity

From equilibrium the impulse is equal to the momentum, therefore

I = p

Ft = m(v_1-v_2)

PART A) Since the body ends at rest, we have the final speed is zero, so the momentum would be

p=m(v_1-v_2)

p = 75*0.15

p = 1125Kg\cdot m/s

Therefore the magnitude of the person's impulse is 1125Kg.m/s

PART B) From the equation obtained previously we have that the Force would be:

Ft = m(v_1-v_2)

F(0.025)= 1125

F= 45000N

Therefore the magnitude of the average force the airbag exerts on the person is 45000N

6 0
3 years ago
A charged object traveling 7 m in a uniform electric field of 5 N/C experiences a 4 J increase in Kinetic Energy.
Travka [436]

To solve this problem it is necessary to apply the principles of conservation of Energy in order to obtain the final work done.

The electric field in terms of the Force can be expressed as

E = \frac{F}{q} \rightarrow F=Eq

Where,

F = Force

E= Electric Field

q = Charge

Puesto que el trabajo realizado es equivalente al cambio en la energía cinetica entonces tenemos que

KE = W

KE = F*d

In the First Case,

4 = (qE)d\\q = \frac{4}{Ed}\\q = \frac{4}{5*7}\\q = 0.1142C

In Second Case,

KE = q E'd

KE = (0.1142)(40)(7)

KE = 31.976J

The total energy change would be subject to,

\Delta KE = 31.976-4

\Delta KE = 27.976J

Therefore the Kinetic Energy change of the charged object is 27.976J

3 0
3 years ago
Which situation is contrary to Newton’s first law of motion?
Bond [772]

Answer:

An object at rest stays at rest as long as unbalanced forces act on it.

Explanation:

Inertia can be defined as the tendency of an object or a body to continue in its state of motion or remain at rest unless acted upon by an external force.

In physics, Sir Isaac Newton's First Law of Motion is known as Law of Inertia and it states that, an object or a physical body in motion will continue in its state of motion at continuous velocity (the same speed and direction) or, if at rest, will remain at rest unless acted upon by an external force.

The inertia of a physical object such as a truck is greatly dependent or influenced by its mass; the higher the quantity of matter in a truck, the greater will be its tendency to continuously remain at rest.

Hence, the situation which is contrary to Newton’s first law of motion is that, an object at rest stays at rest as long as unbalanced forces act on it.

According to Newton’s first law of motion, an object at rest stays at rest as long as unbalanced forces do not act on it.

4 0
3 years ago
A baseball player hits a baseball with a bat. The mass of the ball is 0.25 kg. The ball accelerates at 200 m/s2. What is the for
Brums [2.3K]

Answer:

50N

Explanation:

Force (N) = mass (kg) × acceleration (m/s²)

0.25kg times 200m/s² = 50N

6 0
3 years ago
You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 6 m at your feet, then a
Zigmanuir [339]

Answer:

4.5 m/s

Explanation:

The rock must barely clear the shelf below, this means that the horizontal distance covered must be

d_x = 5 m

while the vertical distance covered must be

d_y = 6 m

The rock is thrown horizontally with velocity v_x, so we can rewrite the horizontal distance as

d_x = v_x t

where t is the time of flight. Re-arranging the equation,

t=\frac{d_x}{v_x} (1)

The vertical distance covered instead is

d_y = \frac{1}{2}gt^2

where we omit the term ut since the initial vertical velocity is zero. From this equation,

t=\sqrt{\frac{2d_y}{g}} (2)

Equating (1) and (2), we can solve the equation to find v_x:

\frac{d_x}{v_x}=\sqrt{\frac{2d_y}{g}}\\\frac{d_x^2}{v_x^2}=\frac{2d_y}{g}\\v_x = d_x \sqrt{\frac{g}{2d_y}}=5\sqrt{\frac{9.8}{2(6)}}=4.5 m/s

6 0
3 years ago
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