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2 years ago

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A force of 1.35 newtons is required to accelerate a book by 1.5 meters/second2 along a frictionless surface. What is the mass of

the book?
A.
0.45 kilograms
B.
0.90 kilograms
C.
1.35 kilograms
D.
1.5 kilograms
E.
1.8 kilograms

2 answers:

marysya [2.9K]2 years ago

7 0

I believe D is the right answer

malfutka [58]2 years ago

3 0

B. seems like the answer, I'm not sure.

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A 92-kg skier is sliding down a ski slope that makes an angle of 30 degrees above the horizontal direction. The coefficient of k

9966 [12]

Answer:

a = 4.05 m/s²

Explanation:

Known data

m= 92 kg : mass of the skier

θ =30° :angle θ of the ski slope with respect to the horizontal direction

μk= 0.10 : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the skier

W: Weight of the skier : In vertical direction

N : Normal force : perpendicular to the ski slope

f : Friction force: parallel to the ski slope

Calculated of the W

W= m*g

W= 92kg* 9.8 m/s² = 901,6 N

x-y weight components

Wx= Wsin θ= 901,6 N *sin 30° = 450.8 N

Wy= Wcos θ = 901,6 N *cos 30° =780.8 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay ay = 0

N - Wy = 0

N = Wy

N = 780.8 N

Calculated of the f

f = μk* N= 0.10*780.8 N

f = 78.08 N

We apply the formula (1) to calculated acceleration of the skier:

∑Fx = m*ax , ax= a : acceleration of the block

Wx - f = m*a

450.8- 78.08 = ( 92)*a

372.72 = (92)*a

a = (372.72)/ (92)

a = 4.05 m/s²

6 0

2 years ago

A rigid vessel of 0.06 m3 volume contains an ideal gas , CV =2.5R, at 500K and 1 bar.a). if 15000J heat is transferred to the ga

andreev551 [17]

Answer:

Given that

V= 0.06 m³

Cv= 2.5 R= 5/2 R

T₁=500 K

P₁=1 bar

Heat addition = 15000 J

We know that heat addition at constant volume process ( rigid vessel ) given as

Q = n Cv ΔT

We know that

P V = n R T

n=PV/RT

n= (100 x 0.06)(500 x 8.314)

n=1.443 mol

So

Q = n Cv ΔT

15000 = 1.433 x 2.5 x 8.314 ( T₂-500)

T₂=1000.12 K

We know that at constant volume process

P₂/P₁=T₂/T₁

P₂/1 = 1000.21/500

P₂= 2 bar

Entropy change given as

$\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}$

Cp-Cv= R

Cp=7/2 R

Now by putting the values

$\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}$

$\Delta S=1.443\times 3.5\times 8.314\ln \dfrac{1000.21}{500}-1.443\times 8.314\ln \dfrac{2}{1}$

a)ΔS= 20.79 J/K

b)

If the process is adiabatic it means that heat transfer is zero.

So

ΔS= 20.79 J/K

We know that

$\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}$

Process is adiabatic

$\Delta S_{surr}=0$

$\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}$

$\Delta S_{univ}= 20.79 +0$

$\Delta S_{univ}= 20.79$

3 0

2 years ago

An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0 km/hour is blowing southward.

Alex17521 [72]

Answer:

The speed of the plane relative to the ground is 300.79 km/h.

Explanation:

Given that,

Speed of wind = 75.0 km/hr

Speed of plane relative to the air = 310 km/hr

Suppose, determine the speed of the plane relative to the ground

We need to calculate the angle

Using formula of angle

$\sin\theta=\dfrac{v'}{v}$

Where, v'=speed of wind

v= speed of plane

Put the value into the formula

$\sin\theta=\dfrac{75}{310}$

$\theta=\sin^{-1}(\dfrac{75}{310})$

$\theta=14.0^{\circ}$

We need to calculate the resultant speed

Using formula of resultant speed

$\cos\theta=\dfrac{v''}{v}$

Put the value into the formula

$\cos14=\dfrac{v''}{310}$

$v''=\cos14\times310$

$v''=300.79\ km/h$

Hence, The speed of the plane relative to the ground is 300.79 km/h.

6 0

2 years ago

Astronomers have discovered a new planet called "Xandar" beyond the orbit of Pluto (No, not really but I need a fake planet for

Burka [1]

Answer:

m = 1.82E+23 kg

Explanation:

G = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

mG = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

m = 1.82E+23 kg

3 0

2 years ago

An 80 kg skateboarder moving at 3 m/s pushes off with her back foot to move faster. If her velocity increases to 5 m/s, what is

Arturiano [62]

1) The kinetic energy of an object is given by:
$K= \frac{1}{2}mv^2$
where m is the object's mass and v its speed.

By using this equation, we find the initial kinetic energy of the skateboarder:
$K_i= \frac{1}{2}(80 kg)(3 m/s)^2=360 J$
and the final kinetic energy as well:
$K_f= \frac{1}{2}(80 kg)(5 m/s)^2=1000 J$

So, her change in kinetic energy is
$\Delta K=K_f-K_i=1000 J-360 J=640 J$

2) The work-energy theorem states that the work done to increase the speed of an object is equal to the variation of kinetic energy of the object:
$W=\Delta K$
Therefore, the work done by the skateboarder is
$W=\Delta K=640 J$

7 0

2 years ago

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