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2 years ago

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A force is applied to a block sliding along a surface (Figure 2). The magnitude of the force is 15 N, and the horizontal compone

nt of the force is 4.5 N. At what angle (in degrees) above the horizontal is the force directed?

2 answers:

patriot [66]2 years ago

4 0

If I have done my math correctly, your answer is 67.5

denpristay [2]2 years ago

3 0

Explanation:

It is given that,

The magnitude of force applied force on the block is, F = 15 N

The horizontal component of the force, $F_x=4.5\ N$

We need to find the angle (in degrees) above the horizontal is the force directed. The horizontal component of the force is given by :

$F_x=F\ cos\theta$

$4.5=15\ cos\theta$

$\theta=cos^{-1}(\dfrac{4.5}{15})$

$\theta=72.54^{\circ}$

So, the angle above the horizontal force and the surface is 72.54 degrees. Hence, this is the required solution.

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If you found yourself on the see-through side of this one-way mirror what is the best way you could prevent someone on the other

Taya2010 [7]

Answer:

If there is any sheets or padded material in this room you can cover the window, you could turn off all the lights if there is a light switch in the room, you could try to bring a bright flashlight in and shine it into the other room(try to annoy the person watching you so they leave), act really boring and hopefully make the other person lose interest.

Explanation:

(hint) If you actually get in a situation like this place your fingernail against the mirror or glass you think could possibly be a one-way mirror. If there's a gap between your nail and the mirror, it's most likely a genuine mirror :)

7 0

1 year ago

A 6.60-kg block slides with an initial speed of 1.56 m/s up a ramp inclined at an angle of 28.4° with the horizontal. The coeffi

Vlad [161]

Answer:

The distance travel by block before coming to rest is 0.122 m

Explanation:

Given:

Mass of block $m = 6.60$ kg

Initial speed of block $v _{i} = 1.56$ $\frac{m}{s}$

Final speed of block $v_{f} = 0$ $\frac{m}{s}$

Coefficient of kinetic friction $\mu _{k} = 0.62$

Ramp inclined at angle $\theta =$ 28.4°

Using conservation of energy,

Work done by frictional force is equal to change in energy,

$\mu _{k} mgd \cos 28.4 = \Delta K - \Delta U$

Where $\Delta U = mg d\sin 28.4$

$\mu _{k} mgd \cos 28.4 = \frac{1}{2}mv_{i} ^{2} - mgd\sin 28.4$

$\mu _{k} mgd \cos 28.4 +mgd\sin 28.4 = \frac{1}{2}mv_{i} ^{2}$

$d(6.60 \times 9.8 \times 0.62 \times 0.879 + 6.60 \times 9.8 \times 0.475) = \frac{1}{2} \times 6.60 \times (1.56)^{2}$

$d = 0.122$ m

Therefore, the distance travel by block before coming to rest is 0.122 m

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2 years ago

B. A hydraulic jack has a ram of 20 cm diameter and a plunger of 3 cm diameter. It is used for lifting a weight of 3 tons. Find

lozanna [386]

Answer:

option (b)

Explanation:

According to the Pascal's law

F / A = f / a

Where, F is the force on ram, A be the area of ram, f be the force on plunger and a be the area of plunger.

Diameter of ram, D = 20 cm, R = 20 / 2 = 10 cm

A = π R^2 = π x 100 cm^2

F = 3 tons = 3000 kgf

diameter of plunger, d = 3 cm, r = 1.5 cm

a = π x 2.25 cm^2

Use Pascal's law

3000 / π x 100 = f / π x 2.25

f = 67.5 Kgf

4 0

2 years ago

The magnetic field of an electromagnetic wave in a vacuum is Bz =(2.4μT)sin((1.05×107)x−ωt), where x is in m and t is in s. You

tatiyna

Answer:

Explanation:

Given

$B_z=(2.4\mu T)\sin (1.05\times 10^7x-\omega t)$

Em wave is in the form of

$B=B_0\sin (kx-\omega t)$

where $\omega =frequency\ of\ oscillation$

$k=wave\ constant$

$B_0=Maximum\ value\ of\ Magnetic\ Field$

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$k=1.05\times 10^7 m^{-1}$

Wavelength of wave $\lambda =\frac{2\pi }{k}$

$\lambda =\frac{2\pi }{1.05\times 10^7}$

$\lambda =5.98\times 10^{-7} m$

7 0

2 years ago

A gas is compressed from 600 cm3 to 200cm3 at a constant pressure of 400 kpa. at the same time, 100 j of heat energy is transfer

Mekhanik [1.2K]

The initial volume of the gas is
$V_i = 600 cm^3$
while its final volume is
$V_f = 200 cm^3$
so its variation of volume is
$\Delta V = V_f - V-i = 200 cm^3 - 600 cm^3 = -400 cm^3 = -400 \cdot 10^{-6} m^3$

The pressure is constant, and it is
$p=400 kPa = 400 \cdot 10^3 Pa$

Therefore the work done by the gas is
$W=p\Delta V = (400 \cdot 10^3 Pa)(-400 \cdot 10^{-6} m^3)=-160 J$
where the negative sign means the work is done by the surrounding on the gas.

The heat energy given to the gas is
$Q=+100 J$

And the change in internal energy of the gas can be found by using the first law of thermodynamics:
$\Delta U = Q-W = 100 J - (-160 J)=+260 J$
where the positive sign means the internal energy of the gas has increased.

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2 years ago