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Doss [256]
2 years ago
5

A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a

distance of 4 m. Given that the vertical component of the pulling force is 12 N, calculate the work done by the force in moving the crate.
Physics
1 answer:
Rasek [7]2 years ago
7 0

Answer:

W=173.48J

Explanation:

information we know:

Total force: F=45N

Weight: w=100N

distance: 4m

vertical component of the force: F_{y}=12N

-------------

In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: F_{x}=Fcos\theta

vertical component: F_{y}=Fsen\theta

but from the given information we know that F_{y}=12N

so, equation these two F_{y}=Fsen\theta and F_{y}=12N

Fsen\theta =12N

and we know the force F=45N, thus:

45sen\theta=12

now we clear for \theta

sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:

F_{x}=Fcos\theta

F_{x}=45cos(15.466)\\F_{x}=43.37N

whith this horizontal component we calculate the work to move the crate a distance of 4 m:

W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J

the work done is W=173.48J

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sweet [91]
The magnitude of the force<span> a 1.5 x 10-3 C charge exerts on a 3.2 x 10-4 C charge located 1.5 m away is 1920 Newtons. The formula used to solve this problem is:

F = kq1q2/r^2

where:
F = Electric force, Newtons
k = Coulomb's constant, 9x10^9 Nm^2/C^2
q1 = point charge 1, C
q2 = point charge 2, C
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Using direct substitution, the force F is determined to be 1920 Newtons.</span>
7 0
2 years ago
A worker stands still on a roof sloped at an angle of 35° above the horizontal. He is prevented from slipping by static friction
aleksley [76]

Answer:

99.63 kg

Explanation:

From the force diagram

N = normal force on the worker from the surface of the roof

f = static frictional force = 560 N

θ = angle of the slope = 35

m = mass of the worker

W = weight of the worker = mg

W Cosθ = Component of the weight of worker perpendicular to the surface of roof

W Sinθ = Component of the weight of worker parallel to the surface of roof

From the force diagram, for the worker not to slip, force equation must be

W Sinθ = f

mg Sinθ = f

m (9.8) Sin35 = 560

m = 99.63 kg

5 0
2 years ago
During a hurricane, the atmospheric pressure inside a house may blow off the roof because of the reduced pressure outside. If ai
m_a_m_a [10]

Answer:

817.5 Pa

Explanation:

From Bernoulli's equation, considering thst there is no height difference then

P1+½d(v1)²=P2+½d(v2)²

P1-P2=½d(v2²-v1²)

∆P=½d(v2²-v1²)

Where P represent pressure, d is density and v is velocity. Subscripts 1 and 2 represent inside and outside. ∆P is tge change in pressure

Given the speed at roof top as 128 km/h, we convert it to m/s as follows

128*1000/3600=35.555555555555=35.56 m/s

Velocity at the bottom of roof is 0 m/s

Density is given as 1.293 kg/m³

∆P=½*1.293*(35.56²-0)=817.5 Pa

5 0
2 years ago
A tennis player serves a tennis ball such that it is moving horizontally when it leaves the racquet. When the ball travels a hor
nalin [4]

Answer:

u_x=38.13\ m/s

Explanation:

Given that initially ball moves in the horizontal direction ,it means that the velocity in the vertical direction is zero.

Horizontal distance = 13 m

Vertical distance = 57 cm

Lets take time to cover 57 cm distance in vertical direction is t.

We know that g is the constant acceleration in the vertical direction so we can apply the equation of motion in the vertical direction.

S=u_yt+\dfrac{1}{2}gt^2

Here u_y=0

S= 57 cm

0.57=0\times t+\dfrac{1}{2}\times 9.81\times t^2

t=0.34 s

Now in the horizontal direction

x=u_xt

Here x=13 m

t= 0.34 s

So

13=u_x\times 0.34

u_x=38.13\ m/s

So the initial speed of ball is 38.13 m/s.

7 0
2 years ago
Capillary action in trees can transport water from the roots to the tree's branches. The capillaries (Xyelem) in a certain tree
lesantik [10]

Answer:

h=14.2857\,m

Explanation:

Given:

radius of capillary, r=10^{-6}\,m

angle of contact, \theta=0^{\circ}

density of water, \rho=1000\,kg.m^{-3}

surface tension of water, T=0.07 \,N.m^{-1}

height, h = ?

We have the equation for the height of meniscus as:

h=\frac{2T.cos\, \theta}{\rho.g.r}

h=\frac{2\times 0.07\times cos\,0^{\circ}}{1000\times 9.8\times 10^{-6}}

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No, the capillary action alone cannot be the mechanism of water transportation to the top of the trees. Transpiration also creates a suction pressure in the xylem complementary to the ascent of sap and cohesion of water being the other causes of movement of water up in the plants.

6 0
2 years ago
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