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MariettaO [177]

2 years ago

3

A 200-N sled of slides down a frictionless hillside that rises at 37° above the horizontal. What is the magnitude of the force t

hat the hill exerts on the sled parallel to the surface of the hill?

2 answers:

Nezavi [6.7K]2 years ago

8 0

Answer:

120 N

Explanation:

Given the angle of inclination of 200 N sled is 37°.

We need to find the magnitude of the force that the hill exerts on the sled parallel to the surface of the hill.

We can see from the diagram, the force that the hill exerts on the sled parallel to the surface of the hill is

$mg\times sin(37 \°)$

As we have given the weight of the sled was 200 N. So, $mg=200\ N$

Now,

$200\times sin(37 \°)\\=200\times 0.601\\=120\ N$

So, 120 N is the force that sled exerts parallel to the surface of the heal.

vichka [17]2 years ago

7 0

Answer:

120n

Explanation:

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You lower the temperature of a sample of liquid carbon disulfide from 90.3 ∘ C until its volume contracts by 0.507 % of its init

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Answer:

$T_{f}$ = 85.89 ° C

Explanation:

The linear thermal expansion process is given by

ΔL = L α ΔT

For the three-dimensional case, the expression takes the form

ΔV = V β ΔT

Let's apply this equation to our case

ΔV / V = -0.507% = -0.507 10-2

ΔT = (ΔV / V) 1 /β

ΔT = -0.507 10⁻² 1 / 1.15 10⁻³

ΔT = -4.409

$T_{f}$ –T₀ = 4,409

$T_{f}$ = T₀ - 4,409

$T_{f}$ = 90.3-4409

$T_{f}$ = 85.89 ° C

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2 years ago

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A passenger bus is travelling 28.0 m/s to the right when the driver applies the brakes. The bus stops in 5.00 s. What is the acc

MAVERICK [17]

Change in velocity = d(v)
d(v) = v2 - v1 where v1 = initial speed, v2 = final speed
v1 = 28.0 m/s to the right
v2 = 0.00 m/s
d(v) = (0 - 28)m/s = -28 m/s to the right

Change in time = d(t)
d(t) = t2 - t1 where t1 = initial elapsed time, t2 = final elapsed time
t1 = 0.00 s
t2 = 5.00 s
d(t) = (5.00 - 0.00)s = 5.00s

Average acceleration = d(v) / d(t)
(-28.0 m/s) / (5.00 s)
(-28.0 m)/s * 1 / (5.00 s) = -5.60 m/s² to the right

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Molybdenum (Mo) has a BCC crystal structure, an atomic radius of 0.1363 nm, and an atomic weight of 95.94 g/mol. Compute and com

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Answer:

Solid State or Condense matter book

Explanation:

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A magnetic field of 0.080 T is in the y-direction. The velocity of wire segment S has a magnitude of 78 m/s and components of 18

nevsk [136]

Answer:

(I). The motional emf induced between the ends of the segment is $2.88i-0.72k$

(II). The motional emf is zero.

Explanation:

Given that,

Magnetic field = 0.080 T

Velocity of wire segment = 78 m/s

Component in x direction = 18 m/s

Component in y direction = 24 m/s

Component in z direction = 72 m/s

Length = 0.50 m

We need to calculate the motional emf induced between the ends of the segment

Using formula of emf

$\epsilon=(B_{x}\times v_{x})l$

Put the value into he formula

$\epsilon=((0,0.080,0)\times(18,24,72))\times0.50$

$\epsilon=((72\times0.080)i-0j+k(-18\times0.080))\times0.50$

$\epsilon=(5.76i-0j-1.44k)\times0.05$

$\epsilon=2.88i-0.72k$

(II). If the wire segment is parallel to the y-axis then the angle between B and v is zero.

We need to calculate the motional emf

Using formula of emf

$\epsilon =Bv\sin\theta\times l$

Here, $\theta = 0$

$\epsilon =Bv\sin0\times l$

$\epsilon = 0$

Hence, (I). The motional emf induced between the ends of the segment is $2.88i-0.72k$

(II). The motional emf is zero.

6 0

2 years ago

A nonuniform, horizontal bar of mass m is supported by two massless wires against gravity. The left wire makes an angle ϕ1 with

strojnjashka [21]

Answer:

$x=\frac{L}{tan(\phi_1)cot(\phi_2)+1}$

Explanation:

Let 'F₁' and 'F₂' be the forces applied by left and right wires on the bar as shown in the diagram below.

Now, the horizontal and vertical components of these forces are:

$F_{1x} = -F_1cos(\phi_1)\\F_{1y}=F_1sin(\phi_1)\\\\F_{2x}=F_2cos(\phi_2)\\F_{2y}=F_2sin(\phi_2)$

As the system is in equilibrium, the net force in x and y directions is 0 and net torque about any point is also 0. Therefore,

$\sum F_x=0\\F_{1x}=F_{2x}\\F_1cos(\phi_1)=F_2cos(\phi_2)\\\frac{F_1}{F_2}=\frac{cos(\phi_2)}{cos(\phi_1)}-------1$

Now, let us find the net torque about a point 'P' that is just above the center of mass at the upper edge of the bar.

At point 'P', there are no torques exerted by the F₁x and F₂x nor the weight of the bar as they all lie along the axis of rotation.

Therefore, the net torque by the forces $F_{1y}\ and\ F_{2y}$ will be zero. This gives,

$-F_{1y}\times x + F_{2y}(L-x) = 0\\F_{1y}\times x=F_{2y}(L-x)\\x=\frac{F_{2y}(L-x)}{F_{1y}}$

But, $F_{1y}=F_1sin(\phi_1)\ and\ F_{2y}=F_2sin(\phi_2)$

Therefore,

$x=\frac{F_2sin(\phi_2)(L-x)}{F_1sin(\phi_1)}\\\textrm{From equation (1),}\frac{F_2}{F_1}=\frac{cos(\phi_1)}{cos(\phi_2)}\\\therefore x=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}\times (L-x)\\x=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}\times L-\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}\times x\\\\$

$x(1+\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)})=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}L\\x(1+\frac{cos(\phi_1)}{sin(\phi_1)}\times \frac{sin(\phi_2}{cos(\phi_2)})=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}L$

We know,

$tan(\phi)=\frac{sin(\phi)}{cos(\phi)}\\\\cot(\phi)=\frac{cos(\phi)}{sin(\phi)}$

∴ $x=\frac{L}{tan(\phi_1)cot(\phi_2)+1}$

6 0

2 years ago