Emilio pushes a 100 kg freshman with 200 N of force. How much is the freshman accelerated?

Predict what the MASS would be of something exerting a force of 45N on a spring scale. EXPLAIN how you got

a_sh-v [17]

The mutual forces of gravity between the Earth and an object on or near
its surface are (mass of the object) x (acceleration of gravity on Earth).
These two forces are equal, and we call their strength the "weight" of
the object. It's the number shown on the scale as long as nobody has
their thumb on the scale. In this problem, the force is 45N . (That's
about 10.12 pounds.)

The acceleration of gravity on Earth is about 9.8 meters per second² .

So 45N = (mass in kilograms) x (9.8 meters per second²)

Divide each side by 9.8 : Mass = 45/9.8 = 4.59 kilograms (rounded)

8 0

2 years ago

A 4.5-kg object oscillates on a horizontal spring with an amplitude of 3.8 cm. Its maximum acceleration is 26 m/s2 . Find (a) th

zimovet [89]

Answer:

F = - K X force constant for spring

a = F / m maximum acceleration

F = 4.5 kg * 26 m/s^2 = 117 Newtons

(A) K = 117 N / .038 m = 3079 N/m

ω = (K/M)^1/2 = (117/5)^1/2 = 4.84 / sec

(B) f = ω / 2 pi = 4.84 / 6.28 = .77 /sec

(C) P = 1 / f = 1/.77 = 1.30 sec

3 0

1 year ago

When air mass is caught between two cold fronts the result is a _______ front.

Gre4nikov [31]

A. Occluded
Explanation- At an occluded front, the cold air mass from the cold front meets the cool air that was ahead of the warm front.

3 0

2 years ago

A lightweight vertical spring of force constant k has its lower end mounted on a table. You compress the spring by a distance d,

shusha [124]

Answer:

$v=d\sqrt{\frac{k}{m}}$

Explanation:

In order to solve this problem, we can do an analysis of the energies involved in the system. Basically the addition of the initial potential energy of the spring and the kinetic energy of the mass should be the same as the addition of the final potential energy of the spring and the kinetic energy of the block. So we get the following equation:

$U_{0}+K_{0}=U_{f}+K_{f}$

In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:

$U_{0}=K_{f}$

The initial potential energy of the spring is given by the equation:

$U_{0}=\frac{1}{2}kd^{2}$

the Kinetic energy of the block is then given by the equation:

$K_{f}=\frac{1}{2}mv_{f}^{2}$

so we can now set them both equal to each other, so we get:

$=\frac{1}{2}kd^{2}=\frac{1}{2}mv_{f}^{2}$

This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:

$kd^{2}=mv_{f}^{2}$

so now we can solve this for the final velocity, so we get:

$v=d\sqrt{\frac{k}{m}}$

6 0

2 years ago

2200 kg semi truck driving down the highway has lost control. The truck rolls across the median and into oncoming traffic. The t

serious [3.7K]

Answer:

The semi truck travels at an initial speed of 69.545 meters per second downwards.

Explanation:

In this exercise we see a case of an entirely inellastic collision between the semi truck and the car, which can be described by the following equation derived from Principle of Linear Momentum Conservation: (We assume that velocity oriented northwards is positive)

$m_{S}\cdot v_{S}+m_{C}\cdot v_{C} = (m_{S}+m_{C})\cdot v$ (1)

Where:

$m_{S}$ , $m_{C}$ - Masses of the semi truck and the car, measured in kilograms.

$v_{S}$ , $v_{C}$ - Initial velocities of the semi truck and the car, measured in meters per second.

$v$ - Final speed of the system after collision, measured in meters per second.

If we know that $m_{S} = 2200\,kg$ , $m_{C} = 2000\,kg$ , $v_{C} = 45\,\frac{m}{s}$ and $v = -15\,\frac{m}{s}$ , then the initial velocity of the semi truck is: