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3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Physics

1 answer:

Gelneren [198K]3 years ago

6 0

Work = force x distance (displacement) or the change in Kinetic energy

Carrying a box down a corridor has no external force acting on the box and therefore no work is being done.

Pushing against a locked door doesn't displace the door, so no work is being done.

Suspending a heavy weight with a strong chain requires, no force and actually prevents a force from acting on the chain.

Pulling a trailer up a hill on the other hand requires a person to apply a force to the trailer and displace it up the hill and therefore it requires work.

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Find the mass of an object with force 280 Newton and acceleration as 18 m/s2.

Yuki888 [10]

Answer:

mass of the object is : -

=》15.55 kg

Explanation:

force (f) = mass (M) × Acceleration (A)

=》280 = M × 18

=》M = 280 ÷ 18

=》M = 15.55 kg

8 0

3 years ago

Compare the time period of two simple pendulums of length 4m and 16m at a place.

Vlad1618 [11]

Answer:

the period of the 16 m pendulum is twice the period of the 4 m pendulum

Explanation:

Recall that the period (T) of a pendulum of length (L) is defined as:

$T=2\,\pi\,\sqrt{ \frac{L}{g} }$

where "g" is the local acceleration of gravity.

SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:

$T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2$

therefore the period of the 16 m pendulum is twice the period of the 4 m pendulum.

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3 years ago

You are pulling a 80 kg box with a rope. The force you exert on the box is 90 N at 30 degrees from the ground. What would be the

Feliz [49]

Answer:

Coefficient of dynamic friction= md= 0.09931

Explanation:

To determine the coefficient of dynamic friction we must first match the friction force that is permendicular to the normal force of the block and opposite to the drag force, to the component of the drag force in this same direction. This component on the X axis of the drag force will be:

F= 90N × cos(30°) = 77.9423N

This component on the X axis of the drag force must be equal to the dynamic friction force that is equal to the coefficient of dynamic friction by the normal force of the block weight:

F= md × m × g= 77.9423N

m= mass of the block

md= coefficient of dynamic friction

g= gravity acceleration

F= md × 80kg× 9.81 (m/s²)= 77.9423(kg×m/s²)

md= (77.9423(kg×m/s²) / 784.8 (kg×m/s²)) = 0.09931

7 0

3 years ago

Two point charges totaling 8 μC exert a repulsive force of 0.15 N on one another when separated by 0.5 m. What is the charge on

nexus9112 [7]

Answer:

B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C

Explanation:

From Coulomb's Law the electrostatic repulsive force is given by the following formula:

F = kq₁q₂/r²

where,

F = Repulsive Force = 0.15 N

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q₁ = Magnitude of 1st Charge = ?

q₂ = Magnitude of 2nd Charge = ?

r = Distance between Charges = 0.5 m

Therefore,

0.15 N = (9 x 10⁹ N.m²/C²)q₁q₂/(0.5 m)²

q₁q₂ = (0.15 N)(0.5 m)²/(9 x 10⁹ N.m²/C²)

q₁q₂ = 4.17 x 10⁻¹²

q₁ = (4.17 x 10⁻¹²)/q₂ -------------------- equation (1)

The sum of charges is given as:

q₁ + q₂ = 8 μC

q₁ + q₂ = 8 x 10⁻⁶

using equation (1):

(4.17 x 10⁻¹²)/q₂ + q₂ = 8 x 10⁻⁶

(4.17 x 10⁻¹²) + q₂² = 8 x 10⁻⁶ q₂

q₂² - (8 x 10⁻⁶) q₂ + (4.17 x 10⁻¹²) = 0

Solving this quadratic equation:

q₂ = 7.4 x 10⁻⁶ C (OR) q₂ = 0.56 x 10⁻⁶ C

q₂ = 7.4 μC (OR) q₂ = 0.6 μC

Therefore,

q₁ = (4.17 x 10⁻¹² C)/(7.4 x 10⁻⁶ C)

q₁ = 0.6μC

Now, if we solve with q₂ = 0.6 μC, we will get q₁ = 7.4 μC.

Therefore, the correct option will be:

B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C

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