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3 years ago

A mover applies a net force of 28 N to a sofa that has a mass of 70 kg.

Physics

1 answer:

Alinara [238K]3 years ago

8 0

Answer: .4 m/s^2= acceleration

Explanation:

f = m*a

We can rearrange this equation to solve for acceleration. Therefore,

a=f/m

a= 28N/70kg

a= 0.4 m/s^2

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A 6 m/s vector pointing North is added to a 2 m/s vector pointing East. What are the magnitude and direction of the resultant?

irina [24]

Answer:

A + B = C Ax = 2 Ay = 0 Bx = 0 By = 6

Ax + Bx = Cx = 2

Ay + By = Cy = 6

C = (2^2 + 6^2)^1/2 = 6.32

Tan Cy / Cx = 6 / 2 = 3

Cy at 71.6 deg

6 0

2 years ago

A spring has a force constant k, and an object of mass m is suspended from it. The spring is cut in half and the same object is

kenny6666 [7]

Answer:

$f2/f1 = \sqrt{2}$

Explanation:

From frequency of oscillation

$f = 1/2pi *\sqrt{k/m}$

Initially with the suspended string, the above equation is correct for the relation, hence

$f1 = 1/2pi *\sqrt{k/m}$

where k is force constant and m is the mass

When the spring is cut into half, by physics, the force constant will be doubled as they are inversely proportional

$f2 = 1/2pi *\sqrt{2k/m}$

Employing f2/ f1, we have

$f2/f1 = \sqrt{2}$

3 0

2 years ago

A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.1 m/s. The car is a distanc

Igoryamba

It's 61.5m

by solving simultaneously equations
29+d = 6t (bear)
d=4.1t (tourist)

6 0

2 years ago

Read 2 more answers

Problem 12: A 20.0-m tall hollow aluminum flagpole is equivalent in strength to a solid cylinder 4.00 cm in diameter. A strong w

avanturin [10]

Answer:

The deformation in the pole due to force is 0.70 mm.

Explanation:

Given that,

Height = 20.0 m

Diameter = 4.00 cm

Force = 1100 N

We need to calculate the area

Using formula of area

$A=\pi\times r^2$

$A=\pi\times(2.00\times10^{-2})^2$

$A=0.00125\ m^2$

$A=1.25\times10^{-3}\ m^2$

We need to calculate the deformation

Using formula of deformation

$\Delta x=\dfrac{1}{s}(\dfrac{F}{A}\times L)$

Where, s = shear modulus

F = force

l = length

A = area

Put the value into the formula

$\Delta x=\dfrac{1}{2.5\times10^{10}}\times(\dfrac{1100}{1.25\times10^{-3}}\times 20.0)$

$\Delta x=0.000704\ m$

$\Delta x=7.04\times10^{-4}\ m$

$\DElta x=0.70\ mm$

Hence, The deformation in the pole due to force is 0.70 mm.

3 0

3 years ago

You have a mass of 55kg and are riding your frictionless skateboard, which has a mass of 5kg, in a straight line at a speed of 4

Rzqust [24]

Given

m1(mass of the first object): 55 Kg

m2 (mass of the second object): 55 Kg

v1 (velocity of the first object): 4.5 m/s

v2 (velocity of the second object): ?

m3(mass of the object dropped): 2.5 Kg

The law of conservation of momentum states that when two bodies collide with each other, the momentum of the two bodies before the collision is equal to the momentum after the collision. This can be mathemetaically represented as below:

Pa= Pb

Where Pa is the momentum before collision and Pb is the momentum after collision.

Now applying this law for the above problem we get

Momentum before collision= momentum after collision.

Momentum before collision = (m1+m2) x v1 =(55+5)x 4.5 = 270 Kgm/s

Momentum after collision = (m1+m2+m3) x v2 =(55+5+2.5) x v2

Now we know that Momentum before collision= momentum after collision.

Hence we get

270 = 62.5 v2

v2 = 4.32 m/s

7 0

2 years ago