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2 years ago

Based on the information in the table, which two elements are most likely in the same group, and why?

Physics

2 answers:

Gnesinka [82]2 years ago

4 0

Answer:

Explanation:

i got it right in my test

velikii [3]2 years ago

3 0

Answer: bismuth and nitrogen, because they have the same number of valence electrons

Explanation:

Elements are distributed in groups and periods in a periodic table.

Elements that belong to same groups will show similar chemical properties because they have same number of valence electrons.

The number of valence electrons in Bismuth and nitrogen are 5 and thus thus they will show similar chemical properties and thus belong to the same group.

The atomic masses of elements in a group will differ drastically.

The group number has got nothing to be the isolation year.

Thus bismuth and nitrogen belong to same group because they have the same number of valence electrons

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An observer sitting on shore sees a canoe traveling 5.0 m/s east, and a sailboat traveling 15.0 m/s west. What is the velocity o

Dovator [93]

Answer:

Vs/c = 20 [m/s]

Explanation:

This is a problem of relative velocities, as velocity is a vector we can use addition or subtraction of vectors for the solution.

We are asked for the velocity of the sailboat with respect to an observer located in the canoe.

$V_{s/c}=V_{sailboat}-V_{canoe}\\V_{s/c}=15-(-5)\\V_{s/c}=20[m/s]$

Why is the speed of the canoe negative?, it is negative because the canoe moves in the opposite direction to the sailboat.

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2 years ago

How long does it take an automobile traveling 66.7 km/h to become even with a car that is traveling in another lane at 52.7 km/h

tresset_1 [31]

Answer:

The time taken is $t = 32.5 \ s$

Explanation:

From the question we are told that

The speed of first car is $v_1 = 66.7 \ km/h = 18.3 \ m/s$

The speed of second car is $v_2 = 52.7 \ km/h = 14.64 \ m/s$

The initial distance of separation is $d = 119 \ m$

The distance covered by first car is mathematically represented as

$d_t = d_i + d_f$

Here $d_i$ is the initial distance which is 0 m/s

and $d_f$ is the final distance covered which is evaluated as $d_f = v_1 * t$

$d_t = 0 \ m/s + (v_1 * t )$

$d_t = 0 \ m/s + (18.3 * t )$

The distance covered by second car is mathematically represented as

$d_t = d_i + d_f$

Here $d_i$ is the initial distance which is 119 m

and $d_f$ is the final distance covered which is evaluated as $d_f = v_2* t$

$d_t = 119 + 14.64 * t$

Given that the two car are now in the same position we have that

$119 + 14.64 * t = 0 + (18.3 * t )$

$t = 32.5 \ s$

6 0

2 years ago

A 70.0-kg ice hockey goalie, originally at rest, catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. Suppos

brilliants [131]

Answer:

v₂=- 34 .85 m/s

v₁=0.14 m/s

Explanation:

Given that

m₁=70 kg ,u₁=0 m/s

m₂=0.15 kg ,u₂=35 m/s

Given that collision is elastic .We know that for elastic collision

Lets take their final speed is v₁ and v₂

From momentum conservation

m₁u₁+m₂u₂=m₁v₁+m₂v₂

70 x 0+ 0.15 x 35 = 70 x v₁ + 0.15 x v₂

70 x v₁ + 0.15 x v₂=5.25 --------1

v₂-v₁=u₁-u₂ ( e= 1)

v₂-v₁ = -35 --------2

By solving above equations

v₂=- 34 .85 m/s

v₁=0.14 m/s

4 0

2 years ago

If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string

vovikov84 [41]

Complete Question

The speed of a transverse wave on a string of length L and mass m under T is given by the formula

$v=\sqrt{\frac{T}{(m/l)}}$

If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string

Answer:

$(m/l)=\frac{10}{V^2}$

Explanation:

From the question we are told that

Speed of a transverse wave given by

$v=\sqrt{\frac{T}{(m/l)}}$

Maximum Tension is $T=10.0N$

Generally making $(m/l)$ subject from the equation mathematically we have

$v=\sqrt{\frac{T}{(m/l)}}$

$v^2=\frac{T}{(m/l)}$

$(m/l)=\frac{T}{V^2}$

$(m/l)=\frac{10}{V^2}$

Therefore the Linear mass in terms of Velocity is given by

$(m/l)=\frac{10}{V^2}$

8 0

2 years ago

If your brakes give out, why can't you just pull the keys out of the ignition?

Pachacha [2.7K]

There should be a small amount of play in the wheel when the steering is locked. Gently pull the key from the ignition while you slowly jiggle the steering wheel back and forth. If this is the cause of the problem, the key should come out after a little effort.

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2 years ago