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2 years ago

HELP!!! I HAVE NO IDEA!!!!!!!!!!!!!

Physics

1 answer:

LekaFEV [45]2 years ago

8 0

Answer:

Explanation:

The engine should create the same force

1) F = ma

F = M(½a)

therefore

ma = ½Ma

m = ½M

M = 2m

M = 2(0.8)

M = 1.6 kg

2) Unbalanced forces

a. Two people pulling on the same side of a wheelbarrow

c. Two people of opposite sides of a big tire. One pushes the tire and one pulls it with equal force

3) b. The force of friction on the object is different if it is stationary or not

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It depends on what the object is and what planet it is on and what the conditions are.

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2 years ago

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2 years ago

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A circular loop of radius 11.9 cm is placed in a uniform magnetic field. (a) If the field is directed perpendicular to the plane

djverab [1.8K]

Answer:

(a). The strength of the magnetic field is 0.1933 T.

(b). The magnetic flux through the loop is zero.

Explanation:

Given that,

Radius = 11.9 cm

Magnetic flux $\phi=8.60\times10^{-3}\ T m^2$

(a). We need to calculate the strength of the magnetic field

Using formula of magnetic flux

$\phi=BA\cos\theta$

$\phi=BA\cos0$

$\phi=BA$

$B=\dfrac{\phi}{A}$

$B=\dfrac{\phi}{\pi r^2}$

Put the value into the formula

$B=\dfrac{8.60\times10^{-3}}{\pi\times(11.9\times10^{-2})^2}$

$B=0.1933\ T$

(b). If the magnetic field is directed parallel to the plane of the loop,

We need to calculate the magnetic flux through the loop

Using formula of flux

$\phi=BA\cos\theta$

Here, $\theta=90^{\circ}$

$\phi=BA\cos90$

$\phi=0$

Hence, (a). The strength of the magnetic field is 0.1933 T.

(b). The magnetic flux through the loop is zero.

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3 years ago

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In a heat engine if 1000 j of heat enters the system the piston does 500 j of work, what is the final internal energy of the sys

son4ous [18]

Answer:

2500 J

Explanation:

We can solve the problem by using the first law of thermodynamics:

$\Delta U =U_f - U_i =Q-W$

where

Uf is the final internal energy of the system

Ui is the initial internal energy

Q is the heat added to the system

W is the work done by the system

In this problem, we have:

Q = +1000 J (heat that enters the system)

W = +500 J (work done by the system)

Ui = 2000 J (initial internal energy)

Using these numbers, we can re-arrange the equation to calculate the final internal energy:

$U_f = U_i + Q-W=2000 J+1000 J-500 J=2500 J$

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3 years ago

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Paraphin [41]

Answer:

1)The position change of almost any manually operated room light switch.

2) Sunlight striking a point on the ground on a partly cloudy and windy day

Explanation:

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2 years ago