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2 years ago

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A crate of oranges with a total mass of 6.7 kg is being pulled across a skating rink (frictionless) with a rope that makes an an

gle of 0.7 rad above the horizontal. John, who measures the motion of the crate, sees that it accelerates at a constant rate of 4.2 m/s2. What is the tension in the rope in N?

2 answers:

AVprozaik [17]2 years ago

5 0

Answer:

The tension in the rope is 36.792 N

Explanation:

Data given:

m = 6.7 kg

θ = 0.7 rad = 40.107°

a = acceleration = 4.2 m/s²

The tension in the rope is:

$T=\frac{ma}{cos\theta } =\frac{6.7*4.2}{cos40.107} =36.792N$

Mama L [17]2 years ago

4 0

Answer:

36.74 N

Explanation:

Given that:

A crate of oranges with a total mass (m) = 6.7 kg

angle θ = 0.7 rad

angle θ = $0.7 * \frac{180}{\pi}$

angle θ = 40°

acceleration = 4.2 m/s²

Given that:

T cosθ = ma

T cos 40° = 6.7 × 4.2

T = $\frac { 6.7 * 4.2}{cos \ 40}$

T = $\frac { 28.14}{0.7660}$

T = 36.74 N

Thus, the tension in the rope = 36.74 N

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Korolek [52]

Answer:

(a) T = 0.015 N

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Explanation:

(a) T = 0.015 N

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$v =f\lambda$

where,

v = speed of wave = ?

f = frequency = 120 Hz

λ = wavelength = 6 cm = 0.06 m

Therefore,

v = (120 Hz)(0.06 m)

v = 7.2 m/s

Now, we will find the linear mass density of the coil:

$\mu = \frac{m}{l}$

where,

μ = linear mass density = ?

m = mass = 1.45 g = 1.45 x 10⁻³ kg

l = length = 5 m

Thereforre,

$\mu = \frac{1.45\ x\ 10^{-3}\ kg}{5\ m}\\\\\mu = 2.9\ x\ 10^{-4}\ kg/m$

Now, for the tension we use the formula:

$v = \sqrt{\frac{T}{\mu}}\\\\7.2\ m/s = \sqrt{\frac{T}{2.9\ x\ 10^{-4}\ kg/m}}\\\\(51.84\ m^2/s^2)(2.9\ x\ 10^{-4}\ kg/m) = T$

<u>T = 0.015 N</u>

<u></u>

(b)

The mass to be hung is:

$T = Mg\\\\M = \frac{T}{g}\\\\M = \frac{0.015\ N}{9.8\ m/s^2}\\\\$

<u>M = 1.53 x 10⁻³ kg = 1.53 g</u>

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What happens if you go through a black hole ?

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Explanation:

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3 years ago

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An archer fires an arrow at a castle 230 m away. The arrow is in flight for 6 seconds, then hits the wall of the castle and stic

Vika [28.1K]

Answer:

(a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

Explanation:

Given that,

Horizontal distance = 230 m

Time t = 6 sec

Vertical distance = 16 m

We need to calculate the horizontal component

Using formula of horizontal component

$R =u\cos\theta t$

Put the value into the formula

$\dfrac{230}{6} = u\cos\theta$

$u\cos\theta=38.33$ .....(I)

We need to calculate the height

Using vertical component

$H=u\sin\theta t-\dfrac{1}{2}gt^2$

Put the value in the equation

$16 =u\sin\theta\times6-\dfrac{1}{2}\times9.8\times6^2$

$u\sin\theta=\dfrac{16+9.8\times18}{6}$

$u\sin\theta=32.06$ .....(II)

Dividing equation (II) and (I)

$\dfrac{u\sin\theta}{u\cos\theta}=\dfrac{32.06}{38.33}$

$\tan\theta=0.8364$

$\theta=\tan^{-1}0.8364$

$\theta=39.90^{\circ}$

(a). We need to calculate the initial speed

Using equation (I)

$u\cos\theta\times t=38.33$

Put the value into the formula

$u =\dfrac{230}{6\times\cos39.90}$

$u=49.96\ m/s$

(b). We have already calculate the angle.

Hence, (a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

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3 years ago

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vesna_86 [32]

Answer:

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Explanation:

Given that,

The magnitude of magnetic field, B = 2.21

We need to find the magnitude of the electric field. Let it is E. So,

$\dfrac{E}{B}=c\\\\E=Bc$

Put all the values,

$E=2.21\times 3\times 10^8\\\\=6.63\times 10^8\ N/C$

So, the magnitude of the electric field is equal to $6.63\times 10^8\ N/C$ .

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2 years ago

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Answer:

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Explanation:

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Therefore, acceleration is $a=4\,\frac{m}{s^2}$

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$y=m\,x+b\\V(t)=4\,t\,+3$

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$V(t)= 4\,t+3\\V(1)=4\,(1)+3\\V(1)=7 \,\frac{m}{s}$

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3 years ago