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3 years ago

What is the acceleration of the object? m/s2

Physics

1 answer:

Molodets [167]3 years ago

8 0

Answer:use

Explanation:google

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The first law of motion applies to what?

lana66690 [7]

First law of motion- sometimes referred to as the law of inertia. An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

3 0

3 years ago

Help please physics !!

malfutka [58]

Answer:

Option A. 1 bar = 1 atm

Explanation:

Pressure has various units of measurement. Each unit of measurement can be converted to other units of measurement. For example:

1 atm = 1 bar

1 atm = 760 mmHg

1 atm = 760 torr

1 atm = 1×10⁵ N/m²

1 atm = 1×10⁵ Pa

With the above conversion scale we can convert from one unit to the other.

Considering the question given above, it is evident from the coversion scale illustrated above that only option A is correct.

Thus,

1 bar = 1 atm

5 0

2 years ago

A 1000 kg car is moving on flat road at a constant speed of 25 m/s. What is the magnitude of average not force, that is required

Shkiper50 [21]

Answer:-600N

Explanation:

Mass(m)=1000kg

Initial velocity(u)=25m/s

Final velocity(v)=12.5m/s

Time(t)=22 seconds

v=u+axt

12.5=25+ax22

collect like terms

12.5-25=22a

-12.5=22a

Divide both sides by 22

-12.5/22=22a/22

-0.6=a

Acceleration=-0.6

Force=mass x acceleration

Force=1000 x -0.6

Force=-600

4 0

3 years ago

Which statement is true about the relationship between carrying capacity and population size?

TiliK225 [7]

Carry capacity determines maximum population size

8 0

3 years ago

A single-turn plane loop of wire with a cross-sectional area 200 cm2 is perpendicular to a magnetic field that increases uniform

BartSMP [9]

Answer:

Induced current, $I=2.87\times 10^{-3}\ A$

Explanation:

Given that,

Area of cross section of the wire, $A=200\ cm^2=0.02\ m^2$

Time, t = 2.2 s

Initial magnetic field, $B_i=0.2\ T$

Final magnetic field, $B_f=2.8\ T$

Resistance of the coil, R = 8 ohms

The expression for the induced emf is given by :

$\epsilon=-\dfrac{d\phi}{dt}$

$\phi$ = magnetic flux

$\epsilon=-\dfrac{d(BA)}{dt}$

$\epsilon=A\dfrac{d(B)}{dt}$

$\epsilon=A\dfrac{B_f-B_i}{t}$

$\epsilon=0.02\times \dfrac{2.8-0.2}{2.2}$

$\epsilon=-0.023\ volts$

So, the induced emf in the loop is 0.023 volts. The induced current can be calculated using Ohm's law as :

$\epsilon= IR$

$I=\dfrac{\epsilon}{R}$

$I=\dfrac{0.023}{8}$

$I=2.87\times 10^{-3}\ A$

So, the magnitude of the induced current in the loop of wire is $2.87\times 10^{-3}\ A$ . Hence, this is the required solution.

7 0

3 years ago