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2 years ago

Which formula is used to find an object's acceleration?

Physics

1 answer:

kompoz [17]2 years ago

3 0

Answer:

${ \rm{Q = \frac{ \triangle \: v}{ \triangle \: t} }} \\$

Explanation:

From definition of acceleration;

${ \tt{Acceleration = \frac{ \delta \: velocity}{time} }} \\$

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Two resistances of 100ohms and 0 ohms were connected in parallel. What is the overall resistance

Inga [223]

Answer:

20 ohms,100*2/10=20

Explanation:

since it as to be in parallel arrangement and connected in side by side due to same p.d in a cross each resistor

6 0

3 years ago

A 0.075 kg ball in a kinetic sculpture is raised 1.33 m above the ground by a motorized vertical conveyor belt. A constant frict

irinina [24]

Answer : The total work done in raising the ball is, 0.98 J

Explanation : Given,

Mass of the ball = 0.075 kg

Height raised of the ball = 1.33 m

As we know that the object is moving with the constant velocity, that means the work done against the gravity will be the net-work done.

So, the work done will be:

$w=mgh$

where,

w = work done

m = mass of ball

h = height of ball

g = acceleration due of gravity = $9.8m/s^2$

Now put all the given values in the above formula, we get:

$w=(0.075kg)\times (9.8m/s^2)\times (1.33m)$

$w=0.97755J=0.98J$

Thus, the total work done in raising the ball is, 0.98 J

5 0

3 years ago

A homeowner is trying to move a stubborn rock from his yard. By using a a metal rod as a lever arm and a fulcrum (or pivot point

Nonamiya [84]

Answer:

1.52 m

Explanation:

We are given that

Maximum force=F=679 N

Mass of rock ,m=385 kg

Distance,d=0.233 m

We have to find the minimum total length L of the rod required to move the rock.

Torque on rock= $T_1=mgd=385\times 9.8\times 0.233=879.11 N$

Where $g=9.8m/s^2$

Torque on man= $T_2=Force\times distance=(L-d)\times 679=(L-0.233)\times 679$

$T_1=T_2$

$(L-0.233)\times 679=879.11$

$L-0.233=\frac{879.11}{679}=1.29$

$L=1.29+0.233=1.523\approx 1.52 m$

Hence, the minimum total length of rod=1.52 m

3 0

3 years ago

A horizontal insulating rod of length 11.8-cm and charge 19 nC is in a plane with a long straight vertical uniform line charge.

SCORPION-xisa [38]

Answer:

11.962337 × 10^-4 N

Explanation:

Given the following :

Length L = 11.8

Charge = 29nC = 29 × 10^-9 C

Linear charge density λ = 1.4 × 10^-7 C/m

Radius (r) = 2cm = 2/100 = 0.02 m

Using the relation:

E = 2kλ/r ; F =qE

F = 2kλq/L × ∫dr/r

F = 2*k*q*λ/L × (In(0.02 + L) - In(0.02))

2*k*q*λ/L = [2 × (9 * 10^9) * (29 * 10^9) * (1.4 * 10^-7)]/ 0.118] = 6193.2203 × 10^(9 - 9 - 7) = 6193.2203 × 10^-7 = 6.1932203 × 10^-4

In(0.02 + 0.118) - In(0.02) = In(0.138) - In(0.02) = 1.9315214

Hence,

(6.1932203 × 10^-4) × 1.9315214 = 11.962337 × 10^-4 N

3 0

4 years ago

A force of 100N is applied to an area of 100mm².what is the pressure exerted on the area in N/m².

Dmitry_Shevchenko [17]

Answer:

P = 1000000[Pa] = 1000 [kPa]

Explanation:

To solve this problem we must use the definition of pressure, which is equal to the relationship of force over area.

$P=F/A$

where:

P = pressure [Pa] (units of pascals)

F = force = 100 [N]

A = area = 100 [mm²]

But first we must convert the units from square millimeters to square meters.

$A=100[mm^{2}]*\frac{1^{2} m^{2} }{1000^{2}mm^{2} } =0.0001[m^{2} ]$

Now replacing:

$P=100/0.0001\\P=1000000[Pa]$

3 0

3 years ago