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Mashcka [7]
3 years ago
12

A 22 kg sled is pushed for 5.2 m with a horizontal force of 20 N, starting from rest. ignore friction. find the final speed of t

he sled.
2.2 m/s
4.7 m/s
9.5 m/s
3.1 m/s
Physics
1 answer:
maksim [4K]3 years ago
5 0
We first calculate the friction on the sled using:
F = ma
a = 20 / 22
a = 0.91 m/s²

Now we use the formula:
2as = v² - u²
v = √(2as + u²)
v = √[2(0.91)(5.2) + 0]
v = 3.08 m/s
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when an object is charged by contact, what kind of charge does the object have compared with the charge on the on the object giv
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Explanation:

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A steel ball of mass 0.500 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released whe
Liula [17]

Answer:

a) v₁fin = 3.7059 m/s   (→)

b) v₂fin = 1.0588 m/s     (→)

Explanation:

a) Given

m₁ = 0.5 Kg

L = 70 cm = 0.7 m

v₁in = 0 m/s   ⇒  Kin = 0 J

v₁fin = ?

h<em>in </em>= L = 0.7 m

h<em>fin </em>= 0 m   ⇒    U<em>fin</em> = 0 J

The speed of the ball before the collision can be obtained as follows

Einitial = Efinal

⇒ Kin + Uin = Kfin + Ufin

⇒ 0 + m*g*h<em>in</em> = 0.5*m*v₁fin² + 0

⇒ v₁fin = √(2*g*h<em>in</em>) = √(2*(9.81 m/s²)*(0.70 m))

⇒ v₁fin = 3.7059 m/s   (→)

b)  Given

m₁ = 0.5 Kg

m₂ = 3.0 Kg

v₁ = 3.7059 m/s    (→)

v₂ = 0 m/s

v₂fin = ?

The speed of the block just after the collision can be obtained using the equation

v₂fin = 2*m₁*v₁ / (m₁ + m₂)

⇒  v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)

⇒  v₂fin = 1.0588 m/s     (→)

7 0
3 years ago
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7 0
3 years ago
Two wires A and B with circular cross-section are made of the same metal and have equal lengths, but the resistance of wire A is
Nesterboy [21]

Answer:

r₁/r₂ = 1/2 = 0.5

Explanation:

The resistance of a wire is given by the following formula:

R = ρL/A

where,

R = Resistance of wire

ρ = resistivity of the material of wire

L = Length of wire

A = Cross-sectional area of wire = πr²

r = radius of wire

Therefore,

R = ρL/πr²

<u>FOR WIRE A</u>:

R₁ = ρ₁L₁/πr₁²   -------- equation 1

<u>FOR WIRE B</u>:

R₂ = ρ₂L₂/πr₂²   -------- equation 2

It is given that resistance of wire A is four times greater than the resistance of wire B.

R₁ = 4 R₂

using values from equation 1 and equation 2:

ρ₁L₁/πr₁² = 4ρ₂L₂/πr₂²

since, the material and length of both wires are same.

ρ₁ = ρ₂ = ρ

L₁ = L₂ = L

Therefore,

ρL/πr₁² = 4ρL/πr₂²

1/r₁² = 4/r₂²

r₁²/r₂² = 1/4

taking square root on both sides:

<u>r₁/r₂ = 1/2 = 0.5</u>

6 0
3 years ago
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