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2 years ago

5. Annie drags her little red wagon with a mass of 5.00 kg, up a hill that has an angle of

Physics

1 answer:

maxonik [38]2 years ago

6 0

Answer: $25.08\ m/s^2$

Explanation:

Given

mass of wagon m=5 kg

elevation

$\theta =30.9^{\circ}\\F=150\ N$

Sin component of weight will oppose the applied force therefore we can write

$F-W\sin \theta=ma\\where\\W=weight(mg)\\a=acceleration$

$150-5\times 9.8\times \sin (30.09)^{\circ}=5\times a\\125.43=5a\\a=25.08\ m/s^2$

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A 5,400 W motor is used to do work. If the motor is used for 640 s, about how much work could it do?

Gemiola [76]

Answer:

3,500,000 J

Explanation:

WORK = POWER * TIME

WORK= 5400 * 640

=6456000 J = 3,500,000 J

4 0

3 years ago

5. How does a jack make changing a tire easier?

strojnjashka [21]

Answer: An jack makes changing a tire easier because it lifts up the car to get the tire off of the ground.

Explanation:

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2 years ago

The closest distance a book can be read from a pair of reading eyeglasses (Power = 1.55 dp) is 26.0 cm. What is the near distanc

Mnenie [13.5K]

Answer:

The image distance is 20.0 cm.

Explanation:

Given that,

Power = 1.55 dp

Distance between book to eye = 26.0+3.00=29.0 cm

We need to calculate the focal length

Using formula of focal length

$f = \dfrac{1}{P}$

Put the value into the formula

$f=\dfrac{1}{1.55}$

$f=0.645\ m$

$f=64.5\ cm$

We need to calculate the image distance

Using lens formula

$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{-u}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{64.5}-\dfrac{1}{-29}$

$\dfrac{1}{v}=\dfrac{187}{3741}$

$v=20.0\ cm$

Hence, The image distance is 20.0 cm.

5 0

2 years ago

what is the mass of a cannon ball if a force 2500 N gives the cannon ball an acceleration of 200 m/s squared?

nignag [31]

Using Newton's second law of motion:

F=ma ; [ F = force (N: kgm/s^2);m= mass (kg); a = acceleration (m/s^2)

Given: Find: Formula: Solve for m:

F: 2500N mass:? F=ma Eq.1 m=F/a Eq. 2

a= 200m/s^2

Solution:

Using Eq.2

m= (2500 kgm/s^2)/ (200m/s^2) = 12.5 kg

8 0

3 years ago

If you are in an elevator that speed up then the apparent weight is

Sholpan [36]

Answer:

Fnet - Fg

Explanation:

When an object is in an elevator, its weight varies with respect to the direction of movement of the elevator and the elevators acceleration.

The weight, W, of an object can be expressed as;

W = mg

where m is the object's mass, and g is the acceleration due gravity.

If the object is in an elevator that speed up, an apparent weight would be felt since both mass and elevator are moving against gravitational pull of the earth.

So that,

$W_{net}$ = mg + ma

where: mg is the weight of the object, and ma is the apparent weight.

Apparent weight (ma) = $W_{net}$ - mg

3 0

2 years ago