Ask question

Ask question

All categories

2 years ago

8

7. Imagine you are pushing a 15 kg cart full of 25 kg of bottled water up a 10o ramp. If the coefficient of friction is 0.02, wh

at is the friction force you must overcome to push the cart up the ramp

1 answer:

pentagon [3]2 years ago

4 0

Answer:

The frictional force needed to overcome the cart is 4.83N

Explanation:

The frictional force can be obtained using the following formula:

$F= \mu R$

where $\mu$ is the coefficient of friction = 0.02

R = Normal reaction of the load = $mgcos\theta$ = $25 \times 9.81 \times cos 10$ = $241.52N$

Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.

$F=0.02 \times 241.52N$

F = 4.83 N

Hence, the frictional force needed to overcome the cart is 4.83N

You might be interested in

An airplane flying at 115 m/s due east makes a gradual turn following a circular path to fly south. The turn takes 15 seconds to

ankoles [38]

Answer:

The magnitude of the centripetal acceleration during the turn is $a=12.04\ m/s^2.$

Explanation:

Given :

Speed to the airplane in circular path , v = 115 m/s.

Time taken by plane to turn , t= 15 s.

Also , the plane turns from east to south i.e. quarter of a circle .

Therefore, time taken to complete whole circle is , $T=t\times 4=60\ s.$

Now , Velocity ,

$v=\dfrac{2\pi r}{T}\\\\115=\dfrac{2\times 3.14\times r}{60}\\\\r=1098.73\ m.$

Also , we know :

Centripetal acceleration ,

$a=\dfrac{v^2}{r}$

Putting all values we get :

$a=12.04\ m/s^2.$

Hence , this is the required solution .

5 0

2 years ago

An alloy is made of a material of specific gravity 7.87 and another material of specific gravity 4.50. The alloy of mass 750g ha

julsineya [31]

Answer:

13.9

Explanation:

Apparent weight is the normal force. Sum of the forces on the alloy when it is submerged:

∑F = ma

N + B − W = 0

N + ρVg − mg = 0

6.6 + (0.78 × 1000) V (9.8) − (0.750) (9.8) = 0

V = 9.81×10⁻⁵

If x is the volume of the first material, and y is the volume of the second material, then:

x + y = 9.81×10⁻⁵

(7.87×1000) x + (4.50×1000) y = 0.750

Two equations, two variables. Solve with substitution:

7870 (9.81×10⁻⁵ − y) + 4500 y = 0.750

0.772 − 7870 y + 4500 y = 0.750

0.0222 = 3370 y

y = 6.58×10⁻⁶

x = 9.15×10⁻⁵

The ratio of the volumes is:

x/y = 13.9

8 0

2 years ago

In a 5000 m race, the athletes run 12 1/2 laps; each lap is 400 m.Kara runs the race at a constant pace and finishes in 17.9 min

Ksju [112]

Answer:

No. of laps of Hannah are 7 (approx).

Solution:

According to the question:

The total distance to be covered, D = 5000 m

The distance for each lap, x = 400 m

Time taken by Kara, $t_{K} = 17.9 min = 17.9\times 60 = 1074 s$

Time taken by Hannah, $t_{H} = 15.3 min = 15.3\times 60 = 918 s$

Now, the speed of Kara and Hannah can be calculated respectively as:

$v_{K} = \frac{D}{t_{K}} = \frac{5000}{1074} = 4.65 m/s$

$v_{H} = \frac{D}{t_{H}} = \frac{5000}{918} = 5.45 m/s$

Time taken in each lap is given by:

$(v_{H} - v_{K})t = x$

$(5.45 - 4.65)\times t = 400$

$t = \frac{400}{0.8}$

t = 500 s

So, Distance covered by Hannah in 't' sec is given by:

$d_{H} = v_{H}\times t$

$d_{H} = 5.45\times 500 = 2725 m$

No. of laps taken by Hannah when she passes Kara:

$n_{H} = \frac{d_{H}}{x}$

$n_{H} = \frac{2725}{400} = 6.8$ ≈ 7 laps

3 0

1 year ago

Light from a monochromatic source shines through a double slit onto a screen 5.00 m away. The slits are 0.180 mm apart. The dark

Nitella [24]

Answer:

Wavelength of incident light, $\lambda = 612 nm$

Given:

Distance between slit and screen, x = 5.00 m

slit width, d = 0.180 mm

width of the fringe, $\beta = 1.70 cm = 0.017 m$

Solution:

To calculate the wavelength of the incident light, $\lambda$ :

$\beta = \frac{x\lambda }{d}$

$\lambda = \frac{\beta d}{x}$

$\lambda = \frac{0.017\times 0.180\times 10^{- 3}}{5} = 6.12\times 10^{- 7}m = 612 nm$

$\lambda = 612 nm$

4 0

2 years ago

An infinite sheet of charge is located in the y-z plane at x = 0 and has uniform charge denisity σ1 = 0.51 μC/m2. Another infini

NNADVOKAT [17]

Answer:

E_total = 5.8 10⁴ N /C

Explanation:

In this problem they ask to find the electric field at two points, the electric field is a vector magnitude, so we can find the field for each charged shoah and add them vectorally at the point of interest.

To find the electric field of a charged conductive sheet, we can use the Gauss law,

Ф = E. d S = $q_{int}$ / ε₀

Let us use as a Gaussian surface a small cylinder, with the base parallel to the sheet, the electric field between the sheet and the normal one next to the cylinder has 90º, so its scalar product is zero, the electric field between the sheet and the base has An Angle of 0º, therefore the scalar product is reduced to the algebraic product.

Let's look for the electric field for plate 1

The total flow is the same for each face, as there are two sides of the cylinder

2E A = q_{int} /ε₀

For the internal load we use the concept of surface density

σ = q_{int1} / A

q_{int1} = σ₁ A

Let's replace

2E A = σ₁ A /ε₀

E₁ = σ₁ / 2ε₀

For the other plate we have a field with a similar expression, but of negative sign

E₂ = -σ₂ / 2ε₀

The total field is,

E_total = σ₁ / 2ε₀ + σ₂ / 2ε₀

E_total = (σ₁ + σ₂) / 2ε₀

Let us apply this expression to our case, when placing a sheet without electric charge, a charge is induced for each sheet, the plate 1 that has a positive charge the electric field is protruding to the right and the plate 2 that has a negative charge creates a incoming field, to the right, as the two fields have the same address add

The conductive sheet in the middle pate undergoes an induced load that is created by the other two plates, but because the conductive plate the charges are mobile and are replaced.

E_total = (0.51 +0.52) 10⁻⁶ / 2 8.85 10⁻¹²

E_total = 5.8 10⁴ N /C

Note that the field is independent of the distance between the plates

4 0

2 years ago