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Dimas [21]
2 years ago
14

Newton’s first law of motion was a giant leap forward in scientific thought during Newton’s time. Even today, the idea is someti

mes difficult at first for people to understand.
Which statement is the best example of an object and motion that would make it hard for people to believe Newton’s first law?
a. A rolling ball eventually slows down and comes to a stop.
b. A wagon must be pushed before it begins to move.
c. The heavier the load in a cart, the harder the cart is to turn.
d. A box does not move when pushed equally from opposite sides.
Physics
2 answers:
masha68 [24]2 years ago
6 0

Answer: B

Explanation:

Newton's first law it's law of inertia.

An object at rest will remain at rest (or an object in motion in a straight line at a constant velocity will remain that way) unless it is acted by an unbalanced force.

In A for the ball to slow down and stop, an external force (like friction with air or the floor) needs to be taken in consideration.

In B we can see how we need to make a force on the wagon to make it move.

In C we have an other Newton 's law, force equals mass times acceleration (2nd law)

In D we can see how it does not move because the forces on the box are balanced.

Zanzabum2 years ago
3 0

Answer: a. A rolling ball eventually slows down and comes to a stop.

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How much heat is absorbed by a 75g iron skillet when it’s temperature rises from 5c to 20c?
alexira [117]

Answer: Q=499.5 J  

Explanation:

The heat (thermal energy) absorbed by the iron skillet can be found using the following equation:

Q=m.C.\Delta T   (1)

Where:

Q is the heat  (absorbed)

m=75 g is the mass of the element (iron in this case)

C is the specific heat capacity of the material. In the case of iron is C=0.444\frac{J}{g\°C}

\Delta T=T_{f}-T_{i}=20\°C - 5\°C= 15\°C is the variation in temperature

Knowing this, lets rewrite (1) with these values:

Q=(75 g)(0.444\frac{J}{g\°C})(15\°C)  (2)

Finally:

Q=499.5 J  

6 0
3 years ago
For a short period of time, the frictional driving force acting on the wheels of the 2.5-Mg van is N, where t is in seconds. If
KATRIN_1 [288]

Answer:

15.6m/s

Completed Question;

For a short period of time, the frictional driving force acting on the wheels of the 2.5-Mg van is N= 600t^2 , where t is in seconds. If the van has a speed of 20 km/h when t = 0, determine its speed when t = 5

Explanation:

Mass m = 2500kg

Speed v1 = 20km/h = 20/3.6 m/s = 5.556 m/s

To determine speed v2;

Using the principle of momentum and impulse;

mv1 + ∫₀⁵ F dt = mv2

8 0
3 years ago
Convert the following to gram. a,250 kg b373 mg c,10 quanital,15 ton​
rusak2 [61]

Explanation:

250 KG = 250000 Grams

373 MG = 0.373 Grams

10 Quintals = 1000000 Grams

15 Ton = 15000000 Grams

Please Follow and inbox me

4 0
2 years ago
A squirrel drops an acorn onto the head of an unsuspecting dog. The acorn falls 4.0\,\text m4.0m4, point, 0, start text, m, end
kirill [66]

Answer:

0.903 seconds

Explanation:

To find how many seconds the acorn fall, we can use the formula for distance travelled with constant acceleration:

D = Vo*t + a*t^2/2,

where D is the distance travelled, Vo is the inicial speed, t is the time and a is the acceleration.

In our problem:

Vo = 0,

a = g = 9.81 m/s2,

D = 4 meters.

So, we can solve the equation to find the time:

4 = 0*t +9.81*t^2/2

4.905*t^2 = 4

t^2 = 4/4.905 = 0.8155

t =   0.903 seconds

8 0
3 years ago
Two straight wires are in parallel and carry electrical currents in opposite directions with the same magnitude of 2.0A. The dis
Veronika [31]

Answer:

Explanation:

Two straight wires

Have current in opposite direction

i1=i2=i=2Amps

Distance between two wires

r=5mm=0.005m

Length of one wire is ∞

Length of second wire is 0.3m

Force between the wire,

The force between two parallel currents I1 and I2, separated by a distance r, has a magnitude per unit length given by

F/l = μoi1i2/2πr

F/l=μoi²/2πr

μo=4π×10^-7 H/m

The force is attractive if the currents are in the same direction, repulsive if they are in opposite directions.

F/l = μoi1i2/2πr

F/0.3=4π×10^-7×2²/2π•0.005

F/0.3=1.6×10^-4

Cross multiply

F=1.6×10^-4×0.3

F=4.8×10^-5N

3 0
3 years ago
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