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prohojiy [21]
4 years ago
5

What is some of the evidence for the speed at which glaciers are melting?

Physics
1 answer:
ozzi4 years ago
6 0
That they don’t have any good stuff
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How does the force of gravity between two bodies change when the distance between them doubles? 1. unable to determine; the mass
Rzqust [24]
6. Drop to one quarter of its original value
7 0
3 years ago
A canoe floats in a lake. Inside the canoe is a 25 kg steel solid ball. If the ball is thrown into the lake, does the level of t
Rus_ich [418]

Answer:

Explanation:

volume of ball bearing = 4/3 π r³

= 4/3 x 3.14 x 1.5³

= 14.13 cc.

if D  be the density of steel , weight of the ball bearing

= 14.13 x D x g

For the first case , water will be displaced to keep it floating

weight of displaced water will be equal to weight of steel

weight of displaced water = 14.13 Dg

mass of displaced water = 14.13 D

volume of displaced water = mass / density of water

= 14.13 D / d                             ; (where d is density of water) .

Now when the steel ball bearing is dipped in water , it will displace water equal to its volume only and it will be drowned

its volume = 14 .13 cc

14.13 D / d  >  14.13  ( because D/d is more than one , since D > d )

volume of water displaced in first case is greater

water level will go up higher in first case .

Hence in the second case water level will go down .

Same will happen in case of 25 kg steel .

4 0
3 years ago
The period of a pendulum may be decreased by
jok3333 [9.3K]

Answer:

shortening its length.

8 0
3 years ago
Read 2 more answers
A series-parallel circuit consists of two parallel circuits connected in series across a 45-V source. One parallel branch consis
musickatia [10]

Answer:

The answer to the question is

The current  through R4 = 0.5865 mA

Explanation:

To solve this we list out the known thus

Voltage source = 45-V

Firsrt parallel circiuit has resistances of R1  = 17 kΩ and R2 = 23 kΩ

Second parallel circiuit has resistances of R3  = 45 kΩ and R4 = 55 kΩ

we first find the current flowing in the circuit by finding thr sum of the total resistance in eah parallel circuit

Firsrt parallel circiuit, for circuit in parallel, sum of resistance 1/RT1 = 1/ R1 + 1/R2 = 1/17+1/23 =  0.1023017 therefore RT1 = 1/0.1023017 = 9.775 kΩ

Similarly we have for the second parallel circuit 1/RT2 = 1/R3 + 1/R4

= 1/45 + 1/55 =  0.0404 Hence RT2 = 1/0.0404 = 24.75 kΩ

This means that the 9.775 kΩ and the 24.75 kΩ are in series hence total resistance of the circuit = sum of all resistances in series  

= 9.775 kΩ + 24.75 kΩ = 34.525 kΩ

However current, I is given by V/R = 45-V/34.525 kΩ = 1.303 × 10⁻³ A or 1.303 mA

From the current divider rule, I4 = I × (R4/(R3+R4)

That is the currrent flowing throuhgh R4 = 1.303  × (45/(45+55)) = 0.5865 mA

3 0
4 years ago
A 5.0 V battery storing 43.0 kJ of energy supplies 1.5 A of current to a circuit. How much energy does the battery have left aft
Free_Kalibri [48]

Answer:

B. 16 kJ

Explanation:

Energy = VIt.............. Equation 1

Where V = Voltage, I = Current, t = time

Given: V = 5.0 V, I = 1.5 A, t = 1 h = 3600 s.

Substitute these values into equation 2

E = 5.0(1.5)(3600)

E = 27000 J

E = 27 kJ.

Amount of energy left = 43 kJ - 27 kJ

Amount of energy left = 16 kJ.

Hence the right option is B. 16 kJ

3 0
3 years ago
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