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4 years ago

13

What maximum force do you need to exert on a relaxed spring with a 1.2×104-n/m spring constant to stretch it 6.0 cm from its equ

ilibrium position?

1 answer:

Elena L [17]4 years ago

6 0

Answer:

Maximum force will be equal to 720 N

Explanation:

We have given that spring constant $k=1.2\times 10^4N/m$

Maximum stretch of the spring x = 6 cm = 0.06 m

We have to find the maximum force on the spring

We know that spring force is given by

$F=kx=1.2\times 10^4\times 0.06=720N$

So the maximum force which is necessary to relaxed the spring will be eqaul to 720 N

You might be interested in

The sum of all forces acting on an object

Zina [86]

Answer:

The net force is the vector sum of all the forces that act upon an object. That is to say, the net force is the sum of all the forces, taking into account the fact that a force is a vector and two forces of equal magnitude and opposite direction will cancel each other out.

Explanation:

put in your own words PLEASE, hope it helps

4 0

3 years ago

A 2,700-kg truck runs into the rear of a 1,000-kg car that was stationary. The truck and car are locked together after the colli

Leto [7]

Answer:

6200 J

Explanation:

Momentum is conserved.

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

The car is initially stationary. The truck and car stick together after the collision, so they have the same final velocity. Therefore:

m₁ u₁ = (m₁ + m₂) v

Solving for the truck's initial velocity:

(2700 kg) u = (2700 kg + 1000 kg) (3 m/s)

u = 4.11 m/s

The change in kinetic energy is therefore:

ΔKE = ½ (m₁ + m₂) v² − ½ m₁ u²

ΔKE = ½ (2700 kg + 1000 kg) (3 m/s)² − ½ (2700 kg) (4.11 m/s)²

ΔKE = -6200 J

6200 J of kinetic energy is "lost".

3 0

3 years ago

A car moves in a straight line at 22.0 m/s for 10.0miles, then at 30.0 m/s for another 10.0miles. Calculate the car’s average sp

maw [93]

Answer: 25.38 m/s

Explanation:

We have a straight line where the car travels a total distance $D$ , which is divided into two segments $d=10 miles$ :

$D=d+d=2d$ (1)

Where $d=10mi \frac{1609.34 m}{1 mi}=16093.4 m$

On the other hand, we know speed is defined as:

$S=\frac{d}{t}$ (2)

Where $t$ is the time, which can be isolated from (2):

$t=\frac{d}{S}$ (3)

Now, for the first segment $d=16093.4 m$ the car has a speed $S_{1}=22m/s$ , using equation (3):

$t_{1}=\frac{d}{S_{1}}$ (4)

$t_{1}=\frac{16093.4 m}{22m/s}$ (5)

$t_{1}=731.518 s$ (6) This is the time it takes to travel the first segment

For the second segment $d=16093.4 m$ the car has a speed $S_{1}=30m/s$ , hence:

$t_{2}=\frac{d}{S_{2}}$ (7)

$t_{2}=\frac{16093.4 m}{30m/s}$ (8)

$t_{2}=536.44 s$ (9) This is the time it takes to travel the secons segment

Having these values we can calculate the car's average speed $S_{ave}$ :

$S_{ave}=\frac{d + d}{t_{1} + t_{2}}=\frac{2d}{t_{1} + t_{2}}$ (10)

$S_{ave}=\frac{2(16093.4 m)}{731.518 s +536.44 s}$ (11)

Finally:

$S_{ave}=25.38 m/s$

3 0

3 years ago

Calculate the work against gravity required to build the right circular cone of height 4 m and base of radius 1.2 m out of a lig

Nana76 [90]

Answer:

Work done = 35467.278 J

Explanation:

Given:

Height of the cone = 4m

radius (r) of the cone = 1.2m

Density of the cone = 600kg/m³

Acceleration due to gravity, g = 9.8 m/s²

Now,

The total mass of the cone (m) = Density of the cone × volume of the cone

Volume of the cone = $\frac{1}{3}\pi r^2 h$

thus,

volume of the cone = $\frac{1}{3}\pi 1.2^2\times 4$ = 6.03 m³

therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg

The center of mass for the cone lies at the $\frac{1}{4}$ times the total height

thus,

center of mass lies at, h' = $\frac{1}{4}\times4=1m$

Now, the work gone (W) against gravity is given as:

W = mgh'

W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J

4 0

3 years ago

An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.7

Kazeer [188]

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

$v = \frac{I}{nqA}$

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

$n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}$

Now, we can find the drift speed:

$v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s$

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!

4 0

3 years ago