What are the 3 parts of an atom and give the electric charge of each

Ket [755]

Answer:

Explanation:

The most common formula for average speed is distance traveled divided by time taken. The other formula, if you have the initial and final speed, add the two together, and divide by 2

8 0

3 years ago

Two vehicles, a car and a truck, leave an intersection at the same time. the car heads east at an average speed of 50 miles pe

olya-2409 [2.1K]

The car heads east at an average speed of 50 miles per hour from the intersection point towards East. The truck heads east at an average speed of 60 miles per hour from the intersection point towards South.

The distance of car from the intersection point after t hours is $50t$ .

The distance of truck from the intersection point after t hours is $60t$ .

Since these distances are perpendicular to each other, distance apart d (in miles) at the end of t hours is

$d=\sqrt{(50t)^2+(60t)^2} \\ d=10\sqrt{61} t\\ d=78.1t$

Thus the distance apart is $d=78.1t \;miles$

5 0

3 years ago

How high up is a 5kg object that has 300j of energy

Eddi Din [679]

-- If the object is moving with speed of 10.954 meters per second, then
it has 300J of kinetic energy no matter where it may be located.

-- If the object is 6.118 meters above somewhere, then it has 300J of
gravitational potential energy relative to that place.

7 0

3 years ago

True or false The momentum of a 100kg object traveling at 3 m/s is 300 N

melomori [17]

Answer:

True

Explanation:

Momentum of an object can be defined as the product of its mass and velocity at which it is travelling. With that in mind, momentum = 3*100=300(kg⋅m/s).

One thing to note is the units mentioned. The SI unit of momentum is kg * m/s as it is the product of mass(kilograms) and velocity(meter per second) and not Newton.

5 0

3 years ago

A block is pulled across a flat surface at a constant speed using a force of 50 newtons at an angle of 60 degrees above the hori

vladimir2022 [97]

The magnitude of the friction force is 25 N

Explanation:

To solve this problem, we just have to analyze the forces acting on the block along the horizontal direction. We have:

The horizontal component of the pulling force, $F cos \theta$ , where F = 50 N is the magnitude and $\theta=60^{\circ}$ is the angle between the direction of the force and the horizontal; this force acts in the forward direction
The force of friction, $F_f$ , acting in the backward direction

According to Newton's second law, the net force acting on the block in the horizontal direction must be equal to the product between the mass of the block and its acceleration:

$\sum F_x = ma_x$