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2 years ago

6

A dog jumps 0.80 m to catch a treat. The dog's displacement vector is shown below.

1 answer:

Brrunno [24]2 years ago

6 0

Answer:D

Explanation:

It was right o khan academy

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A spring with spring constant of 26 N/m is stretched 0.22 m from its equilibrium position. How much work must be done to stretch

Evgen [1.6K]

Answer:

1.503 J

Explanation:

Work done in stretching a spring = 1/2ke²

W = 1/2ke²........................... Equation 1

Where W = work done, k = spring constant, e = extension.

Given: k = 26 N/m, e = (0.22+0.12), = 0.34 m.

Substitute into equation 1

W = 1/2(26)(0.34²)

W = 13(0.1156)

W = 1.503 J.

Hence the work done to stretch it an additional 0.12 m = 1.503 J

8 0

2 years ago

A man pushes on a piano with mass 190 kgkg ; it slides at constant velocity down a ramp that is inclined at 18.0 ∘∘ above the ho

creativ13 [48]

Answer:

The magnitude of applied force,parallel to the incline is 575.38 N and parallel to the floor is 605 N.

Explanation:

Given:

Mass of the piano $(m)$ = 190 kg

Inclined angle $(\theta)$ = 18 degree

Considering gravity, $g$ = 9.8 $ms^-^2$

And

Using, $sin(18) =0.30$ and $cos(18)=0.95$

<em>FBD diagram is attached with all the force acting on the floor and and the inclined. </em>

We have to find the magnitude of forces,when the man pushes it parallel to the incline and to the floor.

a.

When the man pushes it parallel to the incline.

Balancing the forces as $\sum F=0$ .

⇒ $F+mgsin(\theta) =0$

⇒ $F=-mgsin(\theta)$

⇒ Here it is negative as the force is acting downward.

⇒ Plugging the values of mass $(m)$ and angle $(\theta)$ .

⇒ $F=190\times 9.8\times sin(18)$

⇒ $F=575.38$ N

b.

When the force is parallel to the floor.

⇒ $Fcos(\theta)=mgsin(\theta)$

⇒ $F=\frac{mgsin(\theta)}{cos(\theta)}$

⇒ Plugging the values.

⇒ $F=\frac{190\times 9.8\times sin(18)}{cos(18)}$

⇒ $F=605$ N

So,

The magnitude of applied force in inclined direction is 575.38 Newton and parallel to the floor is 605 N.

6 0

2 years ago

A block floats partially submerged in a container of liquid. When the entire container is accelerated upward, which of the follo

Nookie1986 [14]

Answer:C

Explanation:

Partially submerged block along with vessel is accelerated upwards .

Initially the block weight is supported by buoyant force such that it is in equilibrium.

when the system start accelerating upwards then the effective gravity will be

g+a where a is the acceleration of the system.

so only net gravity is increased so block will not ascend or descend.

Mathematically

$F_b=weight$

$\rho _L\cdot V\cdot (g+a)=m(g+a)$

where $\rho _L=$ density of liquid

V=volume of object inside the water

7 0

2 years ago

Can a 20 N force and 40 N force ever produce a resultant with magnitude of 27 N?

SVEN [57.7K]

Sure ,Let's find angle between forces

Vectors be A and B and resultant be R

$\\ \sf\longmapsto R^2=A^2+B^2+2ABcos\theta$

$\\ \sf\longmapsto 27^2=20^2+40^2+2(20)(40)cos\theta$

$\\ \sf\longmapsto 729=400+1600+1600cos\theta$

$\\ \sf\longmapsto 729=2000+1600cos\theta$

$\\ \sf\longmapsto 1600cos\theta=-271$

$\\ \sf\longmapsto cos\theta=-0.169$

$\\ \sf\longmapsto \theta=cos^{-1}(-0.169)$

$\\ \sf\longmapsto \theta=80.2°$

7 0

2 years ago

The primary coil of a transformer is connected to a 120 v wall outlet. the secondary coil is connected to a lamp that dissipates

Lyrx [107]

Energy/power is not gained or lost going through a (ideal) transformer.

So the transformer in this problem really doesn't matter. If the lamp is using energy at the rate of 60 watts, then the whole contraption is getting 60 watts of power from the wall outlet.

Power = (voltage) x (current)

60 watts = (120 v) x (current)

Current = (60 watts) / (120 v)

<em>Current = 0.5 Ampere</em>

5 0

2 years ago