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svp [43]
3 years ago
12

A boat crosses a 200m wide river at 3m/s, north relative to water. The river flows at 1m/s as shown. What is the velocity of the

boat as observed by a stationary observer on the river back from the boat departed?
Physics
1 answer:
Xelga [282]3 years ago
3 0

Answer:

The resultant velocity = 3.16 m/s

Explanation:

Since the boat is moving North of the direction of the riverflow, the river would either be flowing westward or Eastward. The two motions form a right angle triangle with the resultant velocity being the hypotenuse of the traingle.

The resultant velocity will be given as ;

R² = B² + r²

Where B is the velocity of the boat and r is the velocity of the river

R² = 3² + 1²

R² = 10

R = √10 = 3.16 m/s

Therefore, the resultant velocity = 3.16 m/s

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A body of volume 100cc immersed completely in water contained in a jar. The weight of water and jar before immersion of the body
BlackZzzverrR [31]
<h2>Answer:</h2>

800gm

<h2>Explanation:</h2>

Archimedes principle states that when an object is immersed in a liquid there is an apparent loss of weight of the object. This apparent loss of weight is also the upthrust experienced by the liquid. The upthrust is equal to the weight of the liquid displaced.

Following from the above statement, when the body of volume 100c.c is immersed in the water contained in the jar, the upthrust experienced is equal to the weight of the water displaced.

<em>Note: In the question, weight is measured just using the mass.</em>

Mass (m) is the product of density (ρ) of liquid (which is water in this case) and volume (v) of body immersed. i.e

m = ρ x v

Where;

ρ = 1 gm/cm³

v = 100c.c = 100cm³

=> m = 1 gm/cm³ x 100cm³

=> m = 100gm

Therefore the weight of water displaced is 100gm

Now, the weight of the water and jar after immersion is the sum of the weight of water and jar before immersion, and the weight of the water displaced. i.e

Weight of water and jar after immersion = 700gm + 100gm = 800gm

8 0
2 years ago
Lorenzo is making a prediction. “I learned that nonmetals increase in reactivity when moving from left to right. So I predict th
nadezda [96]
That prediction is not correct because Xenon is extremely stable; column 18 of the periodic table contains the noble gasses, which are stable because their outer-most energy levels are completely filled. Having the octet (8) of valence electrons means that the element no longer needs to lose or gain electrons to gain stability.

The column 17 elements are unstable because they only have one valence electron short of the stable octet configuration of the noble gasses.
6 0
3 years ago
Read 2 more answers
A 20-cm long solenoid consists of 100 turns of a coil of radius r = 3.0 cm. A current of Io in the coiled wire produces a magnet
Romashka-Z-Leto [24]

Answer:

vi) Double the current in the wire, and double the number of turns in the 20-cm long solenoid

Explanation:

The magnetic field inside the solenoid and the current flowing in the coil of solenoid are related to each other by the following equation

B₀=μ₀nI₀

Where,

B₀ is the magnetic field in the middle of solenoid

n is the number of turns in the coil of solenoid

I₀ is the current flowing in the coil of solenoid

In the above equation, as μ₀ is a constant so the magnetic field will be directly proportional to the number of turns multiplied by the current. So, changing the radius of the coil or length of the coil will have no effect on the magnetic field.

As we have to increase the magnetic field by 4 times, we need to double the current as well as the number of turns as mentioned in the option vi.

3 0
3 years ago
Read 2 more answers
A thin, metallic spherical shell of radius 0.227 m has a total charge of 6.03 × 10 − 6 C placed on it.
KATRIN_1 [288]

Answer:

Explanation:

Given

radius r=0.227 m

Charge on surface Q=6.03\times 10^{-6} C

Point Charge inside sphere q=1.15\times 10^{-6} C

Electric Field at r=0.735 m

Treating Surface charge as Point charge and applying Gauss law

E_{total}A=\frac{q_{enclosed}}{\epsilon _0}

where A=surface area up to distance r

E_{total}=\frac{Q+q}{4\pi r^2}

E_{total}=\frac{6.03\times 10^{-6}+1.15\times 10^{-6}}{4\pi (0.735)^2\times 8.85\times 10^{-12}}

E_{total}=1.194\times 10^{5} N/C

3 0
3 years ago
The student soon loses his balance and falls backwards off the board at a velocity of 1.0 m/s. Assuming momentum is conserved in
Phantasy [73]

Answer:

v2 = 27.3m/s

Explanation:

Assuming forward as positive.

Mass = m1 = 64kg

Let v be the common velocity of the student and the skateboard.

mass of skateboard = m2 = 5.94kg

v = 1.4m/s

Since the skateboard and the student are initially moving together at the same velocity their momentum together is

(m1 + m2)v

Let the final velocity of the student be v1 and the final velocity of the skateboard be v2

v1 = – 1.0m/s (falls backwards that's why the velocity is negative since we are assuming forward as positive)

Then from conservation of momentum, momentum before is equal to momentum after.

(m1 + m2)v = m1v1 + m2v2

m2v2= (m1 + m2)v – m1v1

v2 = ( (m1 + m2)v – m1v1)/m2

v2 = ( (64 + 5.94)×1.4 – 64×(-1.0))/5.94

v2 = ( (64 + 5.94)×1.4 + 64×1.0)/5.94

v2 = 27.3m/s

5 0
3 years ago
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