What is the speed of a bobsled whose distance-time graph indicates that it traveled 117m in 25s

6) If a mass of an object is decreased to half and acting force is reduced by quarter the acceleration of its motion

Zepler [3.9K]

Answer:

Decreases to half.

Explanation:

From the question given above, the following data were obtained:

Initial mass (m₁) = m

Initial force (F₁) = F

Initial acceleration (a₁) =?

Final mass (m₂) = ½m

Final force (F₂) = ¼F

Final acceleration (a₂) =?

Next, we shall determine a₁. This can be obtained as follow:

F₁ = m₁a₁

F = ma₁

Divide both side by m

a₁ = F / m

Next, we shall determine a₂.

F₂ = m₂a₂

¼F = ½ma₂

2F = 4ma₂

Divide both side by 4m

a₂ = 2F / 4m

a₂ = F / 2m

Finally, we shall determine the ratio of a₂ to a₁. This can be obtained as follow:

a₁ = F / m

a₂ = F / 2m

a₂ : a₁ = a₂ / a₁

a₂ / a₁ = F/2m ÷ F/m

a₂ / a₁ = F/2m × m/F

a₂ / a₁ = ½

Cross multiply

a₂ = ½a₁

From the illustrations made above, the acceleration of the car will decrease to half the original acceleration

7 0

3 years ago

When the velocity of an object Changes it is acted upon by a

anygoal [31]

When the velocity of an object changes, it is acted upon by a force

5 0

3 years ago

Please help. 8th grade science

Aneli [31]

Answer:

false

Explanation:

It doesn't the copper wire wouldn't even be pulled by the magnet at all and the electricity would stay inside of the the force of the copper wire

3 0

3 years ago

Two blocks are placed at the ends of a horizontal massless board, as in the drawing. The board is kept from rotating and rests o

Andrew [12]

Answer:

The magnitude of the angular acceleration ∝ = $\frac{rxF}{2.8[tex]r^{2}$ }[/tex]

Explanation:

The angular acceleration ∝ is equal to the torque (radius multiplied by force) divided by the mass times the square of the radius. The magnitude of angular acceleration ∝ will have the equation above but we have to replace the mass in the equation by 2.8kg as stated.

7 0

3 years ago

Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge

NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is $V=\frac{1}{4\pi\epsilon} \frac{q}{d}$

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as- $E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}$

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining two charges where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

5 0

3 years ago