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Lady bird [3.3K]
3 years ago
14

A pendulum consists of a 1.7-kg block hanging on a 1.6-m length string. A 0.01-kg bullet moving with a horizontal velocity of 82

8 m/s strikes, passes through, and emerges from the block (initially at rest) with a horizontal velocity of 340 m/s. To what maximum height above its initial position will the block swing? Write your answer in meters.
Physics
1 answer:
Rzqust [24]3 years ago
6 0

Answer:

0.42 m

Explanation:

mass of pendulum, M = 1.7 kg

Length of pendulum , l = 1.6 m

mass of bullet, m = 0.01 kg

initial velocity of bullet, u = 828 m/s

final velocity of bullet, v = 340 m/s

initial velocity of pendulum, U = 0

Let the final velocity of pendulum is V.

Use conservation of momentum for bullet and the pendulum

m x u + M x U = m x v + M x V

0.01 x 828 + 1.7 x 0 = 0.01 x 340 + 1.7 x V

8.28 + 0 = 3.4 = 1.7 V

V = 2.87 m/s

Now the kinetic energy of the pendulum is converted into potential energy of pendulum and let it raised to a height of h from the initial level.

Use energy conservation

Kinetic energy of the pendulum  = potential energy of the pendulum

0.5 x M x V² = M x g x h

0.5 x 2.87 x 2.87 = 9.8 x h

h = 0.42 m

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The weight of a metal bracelet is measured to be 0.10400 N in air and 0.08400 N when immersed in water. Find its density.
Anna007 [38]

Answer:

The density of the metal is 5200 kg/m³.

Explanation:

Given that,

Weight in air= 0.10400 N

Weight in water = 0.08400 N

We need to calculate the density of metal

Let \rho_{m} be the density of metal and \rho_{w} be the density of water is 1000kg/m³.

V is volume of solid.

The weight of metal in air is

W =0.10400\ N

mg=0.10400

\rho V g=0.10400

Vg=\dfrac{0.10400}{\rho_{m}}.....(I)

The weight of metal in water is

Using buoyancy force

F_{b}=0.10400-0.08400

F_{b}=0.02\ N

We know that,

F_{b}=\rho_{w} V g....(I)

Put the value of F_{b} in equation (I)

\rho_{w} Vg=0.02

Put the value of Vg in equation (II)

\rho_{w}\times\dfrac{0.10400}{\rho_{m}}=0.02

1000\times\dfrac{0.10400}{0.02}=\rho_{m}

\rho_{m}=5200\ kg/m^3

Hence, The density of the metal is 5200 kg/m³.

6 0
3 years ago
An oil film with refractive index 1.48 and thickness 290 nm is floating on water and illuminated with white light at normal inci
VikaD [51]

Answer:

572.3 nm

Explanation:

n_{oil} = refractive index of the oil film = 1.48

t_{oil} = thickness of the oil film = 290 nm

\lambda = wavelength of the dominant color

m = order

Using the equation

2 n_{oil} t_{oil} = (m + 0.5) \lambda

For m = 0

2 (1.48) (290) = (0 + 0.5) \lambda

\lambda = 1716.8 nm

For m = 1

2 (1.48) (290) = (1 + 0.5) \lambda

\lambda = 572.3 nm

For m = 2

2 (1.48) (290) = (2 + 0.5) \lambda

\lambda = 343.4 nm

Hence the dominant color wavelength is 572.3 nm

8 0
3 years ago
A single loop of wire with an area of 0.0900 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendic
erica [24]

Answer:

(a) 0.0171 V

Explanation:

A = 0.09 m^2, dB/dt = 0.190 T/s

(a) According to the law of electromagntic induction

e = dФ / dt

e = A dB / dt

e = 0.09 x 0.190 = 0.0171 V

(b)

as we know

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Answer:

the answer is ture

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