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Westkost [7]
3 years ago
12

An 800 N box is pushed across a level floor for a distance of 5.0 m with a force of 400 N. How much work was done on this box.

Physics
1 answer:
AveGali [126]3 years ago
4 0

Answer: 2000 J.

Explanation: Since work is force*displacement, we just have to multiply the force by the distance: w = f*d = 400 N*5.0 m = 2000 J.

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When an object is in circular motion it is constantly changing its velocity.
EleoNora [17]

Its tangential speed is constant although its velocity is changing. As the object changes direction, it results in a changing of positive and negative signs of the velocity. Although, the magnitude of the velocity (speed) is not changing.

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A metabolic waste of algae that can be recycled for use in cellular respiration is
m_a_m_a [10]

Answer:oxygen

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This oxygen given off is used by organisms for cellular respiration.the mitochondria is the organelle responsible it's utilization in respiration and carbon dioxide is given off.oxygen serves as an electron acceptor in the energy producing process in the mitochondria. It is an important gas for aerobic respiration.

5 0
3 years ago
A 4300-N force from a car's engines produces an acceleration of 3.3 m/s2. What is the mass of the car?
sertanlavr [38]

Answer:The net force acting on the car is 3×103 Newtons

Explanation:

5 0
3 years ago
1. What could be the energy of a flying aeroplane?<br><br> 2. Is work the same as power?
Doss [256]

Explanation:

The total energy of an aircraft flying in the atmosphere can be calculated using equation 1. [2]

E = ½ m v2 + mgh

A Boeing 737-300 has a maximum takeoff weight of 5.65 × 104 kg, a cruise altitude of h = 10,195 m, and cruise speed of 221 m/sec. Inserting these numbers into the above equation, we obtain 7.03 GJ for the energy at cruise conditions. [3] However, the engines mounted onto the wings of the plane are required to provide additional energy per time, power, in order to keep the aircraft flying at a constant altitude and speed

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7 0
3 years ago
A physics student stands on the rim of the canyon and drops a rock. The student measures the time for it to reach the bottom to
zysi [14]

Answer:

50.2 m

Explanation:

We can solve the problem by using the following SUVAT equation for the vertical position of the rock:

y(t)=h+ut+\frac{1}{2}gt^2

where

h is the initial height (the depth of the canyon), taking the bottom of the canyon as reference position

u = 0 is the initial velocity of the rock

t is the time

g=-9.8 m/s^2 is the acceleration of gravity

When the rock reaches the bottom, t = 3.2 s and y = 0. Substituting these numbers and solving for h, we find the depth of the canyon:

h=\frac{1}{2}gt^2 = -\frac{1}{2}(-9.8)(3.2)^2=50.2 m

5 0
3 years ago
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