On the image at right, the two magnets are the same. Which paper clip would be harder to remove?

A fisherman has caught a very large, 5.0kg fish from a dock that is 2.0m above the water. He is using lightweightfishing line th

agasfer [191]

Answer:

t = 2 s

Explanation:

As we know that fish is pulled upwards with uniform maximum acceleration

then we will have

$T - mg = ma$

here we know that maximum possible acceleration of so that string will not break is given as

$T = 54 N$

now we have

$54 - (5 \times 9.8) = 5 a$

$a = 1 m/s^2$

now for such acceleration we can use kinematics

$d = \frac{1}{2}at^2$

$2 = \frac{1}{2}(1) t^2$

t = 2 s

7 0

2 years ago

A block with mass m1 = 4.50 kg and a ball with mass m2 = 7.70 kg are connected by a light string that passes over a frictionless

Allisa [31]

$1.6a = \frac{g(m_2 + m_3 - \mu km_1)}{m_1 + m_2 + m_3} \\ \\ 1.6a(m_1 + m_2 + m_3) = g(m_2 + m_3 - \mu km_1) \\ \\ (1.6a + \mu kg)m_1 + (1.6a - g)m_2 = (g - 1.6a)m_3 \\ \\ m_3 = \frac{1.6a +\mu kg}{g - 1.6a} m_1 - m_2 \\ \\ m_3 = 22.57 kg$

4 0

1 year ago

Deep-sea divers often breathe a mixture of helium and oxygen to avoid getting the "bends" from breathing high-pressure nitrogen.

kvv77 [185]

Answer:

0.69444 m, 0.08152 m, 0.32407 m, 0.03804 m

Explanation:

v = Velocity of sound

f = Frequency

Length of vocal tract is given by

$L=\dfrac{v}{4f}$

At f = 270 Hz v = 750 m/s

$L=\dfrac{750}{4\times 270}\\\Rightarrow L=0.69444\ m$

At f = 2300 Hz v = 750 m/s

$L=\dfrac{750}{4\times 2300}\\\Rightarrow L=0.08152\ m$

At f = 270 Hz v = 350 m/s

$L=\dfrac{350}{4\times 270}\\\Rightarrow L=0.32407\ m$

At f = 2300 Hz v = 350 m/s

$L=\dfrac{350}{4\times 2300}\\\Rightarrow L=0.03804\ m$

3 0

1 year ago

Glider‌ ‌A‌ ‌of‌ ‌mass‌ ‌0.355‌ ‌kg‌ ‌moves‌ ‌along‌ ‌a‌ ‌frictionless‌ ‌air‌ ‌track‌ ‌with‌ ‌a‌ ‌velocity‌ ‌of‌ ‌0.095‌ ‌m/s.‌

NemiM [27]

Answer:

vB' = 0.075[m/s]

Explanation:

We can solve this problem using the principle of linear momentum conservation, which tells us that momentum is preserved before and after the collision.

Now we have to come up with an equation that involves both bodies, before and after the collision. To the left of the equal sign are taken the bodies before the collision and to the right after the collision.

$(m_{A}*v_{A})+(m_{B}*v_{B})=(m_{A}*v_{A'})+(m_{B}*v_{B'})$

where:

mA = 0.355 [kg]

vA = 0.095 [m/s] before the collision

mB = 0.710 [kg]

vB = 0.045 [m/s] before the collision

vA' = 0.035 [m/s] after the collision

vB' [m/s] after the collison.

The signs in the equation remain positive since before and after the collision, both bodies continue to move in the same direction.

$(0.355*0.095)+(0.710*0.045)=(0.355*0.035)+(0.710*v_{B'})\\v_{B'}=0.075[m/s]$

7 0

1 year ago

Three point charges are arranged on a line. Charge q3 = +5.00 nC and is at the origin. Charge q2 = -2.00 nC and is at x = 5.00 c

tatuchka [14]

Answer:

q₁= +0.5nC

Explanation:

Theory of electrical forces

Because the particle q3 is close to three other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

To solve this problem we apply Coulomb's law:

Two point charges (q1, q2) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

o solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

F=K*q₁*q₂/d² Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁,q₂:Charges in Coulombs (C)

d: distance between the charges in meters

Data:

Equivalences

1nC= 10⁻⁹ C

1cm= 10⁻² m

Data

q₃=+5.00 nC =+5* 10⁻⁹ C

q₂= -2.00 nC =-2* 10⁻⁹ C

d₂= 5.00 cm= 5*10⁻² m

d₁= 2.50 cm= 2.5*10⁻² m

k = 8.99*10⁹ N*m²/C²

Calculation of magnitude and sign of q1

Fn₃=0 : net force on q3 equals zero

F₂₃:The force F₂₃ that exerts q₂ on q₃ is attractive because the charges have opposite signs,in direction +x.

F₁₃:The force F₂₃ that exerts q₂ on q₃ must go in the -x direction so that Fn₃ is zero, therefore q₁ must be positive and F₂₃ is repulsive.

We propose the algebraic sum of the forces on q₃

F₂₃ - F₁₃=0

$\frac{k*q_{2} *q_{3} }{d_{2}^{2} } -\frac{k*q_{1} *q_{3} }{d_{1}^{2} }=0$

We eliminate k*q₃ of the equation

$\frac{q_{1} }{d_{1}^{2} } = \frac{q_{2} }{d_{2}^{2} }$

$q_{1} =\frac{q_{2} *d_{1} ^{2} }{d_{2}^{2} }$

$q_{1} =\frac{2*10^{-9}*2.5^{2}*10^{-4} }{5^{2}*10^{-4} }$

q₁= +0.5*10⁻⁹ C

q₁= +0.5nC

4 0

1 year ago