What characteristic describes what happens during a physical change

An object is moving east with a constant speed of 30 m/s for 5 seconds. What is the object’s acceleration?

Shtirlitz [24]

Answer:

6 m/s²

Explanation:

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

Velocity (v) = 30 m/s

Time (t) = 5 s

Acceleration (a) =..?

Acceleration is defined mathematically as:

Acceleration (a) = Velocity (v) /time (t)

a = v /t

With the above formula, we can obtain the acceleration of the object as follow:

Velocity (v) = 30 m/s

Time (t) = 5 s

Acceleration (a) =..?

a= v/t

a= 30/5

a = 6 m/s²

Therefore, the acceleration of the object is 6 m/s² due East.

7 0

3 years ago

Derive a relation between Universal Gravitational Constant (G) and Acceleration due to gravity(g)?

snow_tiger [21]

Suppose earth is a soid sphere which will attract the body towards its centre.So, acc. to law of gravitation force on the body will be,
F=G*m1m2/R^2
but we now that F=ma
and here accleration(a)=accleration due to gravity(g),so
force applied by earth on will also be mg
replace above F in formula by mg and solve,
F=G*mE*m/R^2 ( here mE is mass of earth and m is mass of body)
mg=G*mEm/R^2
so,
g =G*mE/R^2

4 0

3 years ago

Read 2 more answers

A cannonball is fired perfectly horizontally from the top of a 210 m tall cliff. It is fired with an initial velocity of 50 m/s.

pochemuha

Answer:

the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

Explanation:

Given;

height of the cliff, h = 210 m

initial horizontal velocity of the cannonball, Ux = 50 m/s

initial vertical velocity of the cannonball, Uy = 0

The time for the cannonball to reach the ground is calculated as;

$h = u_yt - \frac{1}{2} gt^2\\\\h = 0 - \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 210}{9.8} }\\\\t = 6.55 \ s$

The horizontal distance covered by the cannonball before it hits the ground is calculated as;

$X = U_x \times \ t\\\\X = 50 \times \ 6.55\\\\X = 327.5 \ m$

Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

8 0

3 years ago

A 2.2kg block of ice slides across a rough floor. Its initial velocity is 2.5m/s and its final velocity is 0.50m/s. How much of

kirill [66]

Answer:A 2.2kg block of ice slides across a rough floor. Its initial velocity is 2.5m/s and its final velocity is 0.50m/s. How much of the ice block melted as a result of the work done by friction? (Latent Heat of water is 3.3*10^5J/kg)

Explanation:

7 0

3 years ago

Place the left charge at the 2 cm position and the right one at the 4 cm position. Vary the left and right charge to the values

Vlad1618 [11]

Answer:

Explanation:

Force between two charges can be expressed as follows

F = k q₁ q₂ / d²

q₁ and q₂ are two charges , d is distance between them , k is a constant whose value is 9 x 10⁹

distance between charges is fixed which is 4 -2 = 2 cm = 2 x 10⁻² m

force between 1μC and 4μC

= 9 x 10⁹ x 1 x 4 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 4μC and 1μC

= 9 x 10⁹ x 4 x 1 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 2μC and 2μC

= 9 x 10⁹ x 2 x 2 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 1μC and 2μC

= 9 x 10⁹ x 1 x 2 x 10⁻¹² / ( 2 x 10⁻² )²

= 4.5 x 10 = 45 N

force between 1μC and 8μC

= 9 x 10⁹ x 1 x 8 x 10⁻¹² / ( 2 x 10⁻² )²

= 18 x 10 = 180 N

force between 2μC and 8μC

= 9 x 10⁹ x 1 x 8 x 10⁻¹² / ( 2 x 10⁻² )²

= 36 x 10 = 360 N

Left Charge Right Charge Resulting force(N)

1μC 4μC 90 N

4μC 1μC 90 N

2μC 2μC 90 N

1μC 2μC 45 N

1μC 8μC 180 N

2μC 8μC 360 N

5 0

3 years ago