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devlian [24]
3 years ago
8

Which is a true statement about gender roles?

Physics
1 answer:
lbvjy [14]3 years ago
6 0
I would c is the best answer. As someone who is closed to multiple cultures, I know that gender roles are not necessarily the same across all cultures. Male-female friendships are definitely easier to develop now than in the past. I also know plenty of males who are very in touch with their emotions. Personally, when making friends, I tend to talk to the people with similar interests, simply because then I actually have something to talk about and it’s not awkward.
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A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find th
Igoryamba

A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and the speed when t = 4. f(t) = 12t² + 35 t + 1

Answer:

Velocity = 131 m/s

Speed = 131 m/s

Explanation:

Equation of motion, s = f(t) = 12t² + 35 t + 1

To get velocity of the particle, let us find the first derivative of s

v (t) = ds/dt = 24t + 35

At t = 4

v(4) = 24(4) + 35

v(4) = 131 m/s

Speed is the magnitude of velocity. Since the velocity is already positive, speed is also 131 m/s

5 0
3 years ago
Determine the ratio of the resistivity of pure water to silver?
shutvik [7]
Silver is a very good conductor, this means its resistivity is very low (from table, we can check the precise value, which is \rho_s = 1.6 \cdot 10^{-8} \Omega m).

Pure water, instead, is a very bad conductor, this means its resistivity is very high, of order of k \Omega \cdot m (10^3 \Omega m ). Even without knowing the precise value of the pure water resistivity, we can estimate the ratio between the pure water resistivity and the silver resistivity by comparing the two orders of magnitude:
r= \frac{10^3 \Omega m}{10^{-8} \Omega m}  \sim 10^{11}

Therefore, we can say that the correct answer is
3 \cdot 10^{11} : 1
3 0
3 years ago
During which two time intervals does the particle undergo equal displacement?
san4es73 [151]

Answer:

BC and DE

Explanation:

In the given figure, the velocity time graph is shown. We know that the area under v-t curve gives the displacement of the particle.

Area under AB, d_1=\dfrac{1}{2}\times 2\times 10=10\ m

Area under BC, d_2=\dfrac{1}{2}\times 2\times 4=4\ m

Area under CD, d_3=\dfrac{1}{2}\times 2\times 7=7\ m

Area under DE, d_4=\dfrac{1}{2}\times 2\times 4=4\ m

Area under EF, d_5=\dfrac{1}{2}\times 2\times 3=3\ m

So, form above calculations it is clear that, during BC and DE undergo equal displacement. Hence, the correct option is (c) "BC and DE = 4 meters".

4 0
3 years ago
A toy cannon uses a spring to project a 5.24-g soft rubber ball. The spring is originally compressed by 5.01 cm and has a force
salantis [7]

Answer:

Speed will be equal to 1.40 m/sec

Explanation:

Mass of the rubber ball m = 5.24 kg = 0.00524 kg

Spring is compressed by 5.01 cm

So x = 5.01 cm = 0.0501 m

Spring constant k = 8.08 N/m

Frictional force f = 0.031 N

Distance moved by ball d = 15.8 cm = 0.158 m

Energy gained by spring

KE=\frac{1}{2}kx^2=\frac{1}{2}\times 8.08\times 0.0501^2=0.0101J

Energy lost due to friction

W=Fd=0.031\times 0.158=0.0048J

So remained energy to move the ball = 0.0101 - 0.0048 = 0.0052 J

This energy will be kinetic energy

\frac{1}{2}mv^2=0.0052

\frac{1}{2}\times 0.00524\times v^2=0.0052

v = 1.40 m/sec

7 0
3 years ago
A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod
ivann1987 [24]

Answer:

The inducerd emf is 1.08 V

Solution:

As per the question:

Altitude of the satellite, H = 400 km

Length of the antenna, l = 1.76 m

Magnetic field, B = 8.0\times 10^{- 5}\ T

Now,

When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

e = - l(vec{v}\times \vec{B})

Here, velocity v is perpendicular to the rod

Thus

e = lvB           (1)

For the orbital velocity of the satellite at an altitude, H:

v = \sqrt{\frac{Gm_{E}}{R_{E}} + H}

where

G = Gravitational constant

m_{e} = 5.972\times 10^{24}\ kg = mass of earth

R_{E} = 6371\ km = radius of earth

v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}}{6371\times 1000 + 400\times 1000} = 7670.018\ m/s

Using this value value in eqn (1):

e = 1.76\times 7670.018\times 8.0\times 10^{- 5} = 1.08\ V

5 0
3 years ago
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