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scoray [572]
3 years ago
13

Which statement BEST compares the force of gravity on a school bus on a school day and a weekend? *

Physics
1 answer:
inna [77]3 years ago
7 0

Answer:

(A) The force would be lower on the school day than the weekend.

Explanation:

In  a  school  day  the  bus  is  on  the  road  travelling  with  kids.  But  in  a  weekend  it  is  parked.  When  a  object  stays  still  that  means  the the  force  which  is  working  on  the  earth  by  the  bus  is  equal  to  the  force which  works  on  the  bus  by  the  earth. we  can  understand  it  clearly by,  Newton's second law of motion.  

This pertains to the behavior of objects for which all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object.  so  as  it  is  when  it  is  balanced  it  is  not  moving.

And  on  a  school  day  the  bus  is  moving  so  it  has  a  force  pulling  itself  forward .  so  it  means  that  the  force  which  the  bus  has  is  greater  than  the  gravitational  force.

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PLEASE HELP 40 POINTS AND BRAINLIEST
Vikki [24]

Then the time interval during which the rocket engine provides upward acceleration in 2.1s hope this helps have more photos but can’t put them more in brainliest is appreciated

8 0
3 years ago
Calculate the force of an 8kg object pushing up from the table
zepelin [54]

The correct answer to the question is: 78.4 N.

CALCULATION:

As per the question, the mass of the object is given as m = 8 Kg.

We are asked to calculate the force of 8 Kg object pushing up from the table.

The force pushing up from the table is nothing else than the normal reaction.

The normal reaction of the object is equal to the weight of the object as the object is simply resting on the table.

Hence, normal reaction = Weight

                                               = mg

                                               = 8 × 9.8 N.

                                                = 78.4 N.

Here, g is known as acceleration due to gravity.

Hence, the force of the object pushing up from the table is 78.4 N.


6 0
4 years ago
You can mow an average of 1400 square meters each hour. How many minutes will it take you to mow a lawn with an area of 320000 s
Ksivusya [100]

Answer:

time required to mow lawn is 1274.06 minutes

Explanation:

given data

average mow = 1400 square meters each hour

area = 320000 square feet

to find out

How many minutes will take to mow lawn

solution

we know that here 1 square feet is equal to 0.092903 square meter

so 320000 square feet will be = 0.09203 × 320000 = 29728.9728 square meter

so time required is express as

time required = \frac{distance}{speed}

time required = \frac{29728.9728}{1400}

time required = 21.23 hours

so time required = 21.23 × 60 min = 1274.06

time required to mow lawn is 1274.06 minutes

7 0
3 years ago
A 12 V battery is connected to a circuit with a 4 resistor. What is the current through the resistor?
seraphim [82]
Answer should be 3A
12/4=3
6 0
3 years ago
A slender rod 100.00cm long is used as a meter stick. Twoparallel axes which are perpendicular to the rod are considered.The fir
vredina [299]

Answer:

 I /I_{cm}  = 1.48 ,    The correct answer is d

Explanation:

the moment of inertia is given by

        I = ∫ r² dm

For figures with symmetry it is tabulated.   In the case of a thin variation, the moment of inertia with respect to its center of mass is

       I_{cm} = 1/12 M L2

There is a widely used theorem, which is the parallel axis theorem, where the moment of inertia of any parallel axis, is the moment of mass inertia plus the moment of inertia of the body taken as a particle

      I = I_{cm}  + M D²

Let's put these expressions to our case.

As the bar is one meter long its center of mass that this Enel midpoint corresponds to

      I_{cm}  = 1/12 m L²

      I_{cm}  = 1/12 m 1.00²

      I_{cm}  = 8.33 10⁻² m

Let's use the parallel axes theorem for the axis that passes through x = 30 cm.  The distance from the enrode masses to the axis is

       D = x_{cm}  - 0.30

       D = 0.50 - 0.30 = 0.20 m

       I = I_{cm}  + m D²

       I = 8.33 10⁻² m + m 0.2²

       I = (8.33 10⁻² + ​​4 10⁻²) m    

       I = 12.33 10⁻² m

The relationship between these two moments of inertia

       I /I_{cm}  = 12.333 10⁻² / 8.333 10⁻²

       I /I_{cm}  = 1.48

The correct answer is d

3 0
3 years ago
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