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2 years ago

The table shows the decay in a 59 g sample of bismuth 212 overtime.

Physics

1 answer:

Vikki [24]2 years ago

7 0

Answer: C

14.75g

Explanation:

Given that the half life time = 60.5s

Let No = initial mass = 59g

N = decayed mass

At time t = 0, No = 59g

At time t = 60.5s,

N = No/2 = 59/2

= 29.5g

At time t = 121

N = 29.5/2 = 14.75g

Therefore N = 14.75g

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Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t

yarga [219]

Answer

The rate at which the magnetic field is changing is $[\frac{dB}{dt} ] = 0.000467 T/s$

Explanation

From the question we are told that

The electric field strength is $E = 3.5mV/m = 3.5 *10^{-3} \ V/m$

The radius is $r = 1.5 \ m$

The rate of change of the magnetic field is mathematically represented as

$\frac{d \phi }{dt} = \int\limits^{} {E \cdot dl}$

Where $dl$ is change of a unit length

$\frac{d \phi}{dt} = A * \frac{dB}{dt}$

Where A is the area which is mathematically represented as

$A = \pi r^2$

$E \int\limits^{} { dl} = ( \pi r^2) (\frac{dB}{dt} )$

$E L = ( \pi r^2) (\frac{dB}{dt} )$

where L is the circumference of the circle which is mathematically represented as

$L = 2 \pi r$

$E (2 \pi r ) = (\pi r^2 ) [\frac{dB}{dt} ]$

$E = \frac{r}{2} [\frac{dB}{dt} ]$

$[\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }$

substituting values

$[\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }$

$[\frac{dB}{dt} ] = 0.000467 T/s$

8 0

2 years ago

Imagine that the above hoop is a tire. the coefficient of static friction between rubber and concrete is typically at least 0.9.

Stels [109]

The hoop is attached.

Consider that the friction force is given by:
F = μ·N
 = μ·m·g·cosθ

We also know, considering the forces of the whole system, that:
F = -m·a + m·g·sinθ
and
a = (1/2)·g·sinθ

Therefore:
-(1/2)·m·g·sinθ + m·g·sinθ = μ·m·g·cosθ
(1/2)·m·g·sinθ = μ·m·g·cosθ
μ = (1/2)·m·g·sinθ / m·g·cosθ
 = (1/2)·tanθ

Now, solve for θ:
θ = tan⁻¹(2·μ)
 = tan⁻¹(2·0.9)
 = 61°

Therefore, the maximum angle you could ride down without worrying about skidding is 61°.

5 0

2 years ago

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0

2 years ago

A simple pendulum 0.64m long has a period of 1.2seconds. Calculate the period of a similar pendulum 0.36m long in the same locat

weqwewe [10]

The period of the second pendulum is 0.9 s

Explanation:

The period of a simple pendulum is given by the equation

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration of gravity at the location of the pendulum

For the first pendulum, we have

L = 0.64 m

T = 1.2 s

Therefore we can find the value of g at that location:

$g=(\frac{2\pi}{T})^2 L=(\frac{2\pi}{1.2})^2 (0.64)=17.5 m/s^2$

Now we can find the period of the second pendulum at the same location, which is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where we have

L = 0.36 m (length of the second pendulum)

$g=17.5 m/s^2$

Substituting,

$T=2\pi \sqrt{\frac{0.36}{17.5}}=0.9 s$

#LearnwithBrainly

8 0

2 years ago

An astronomy student, for her PhD, really needs to estimate the age of a cluster of stars. Which of the following would be part

LiRa [457]

Answer:

A. plot an H-R diagram for the stars in the cluster.

Explanation:

A star cluster can be defined as a constellation of stars, due to gravitational force, which has the same origin.

The astronomy student would have to plot an H-R diagram for the stars in the cluster and determine the age of the cluster by observing the turn-off point. The turn-off is majorly as a result of gradual depletion of the source of energy of the star. Thus, it projects off the constellation.

4 0

2 years ago