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Ymorist [56]
4 years ago
9

How many significant figures are in the number 17.95?

Physics
1 answer:
ANTONII [103]4 years ago
6 0

Answer:

4

Explanation:

All of them are significant figures

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the temperature of a 2.0-kg increases by 5*c when 2,000 J of thermal energy are added to the block. What is the specific heat of
nata0808 [166]
To calculate the specific heat capacity of an object or substance, we can use the formula

c = E / m△T

Where
c as the specific heat capacity,
E as the energy applied (assume no heat loss to surroundings),
m as mass and
△T as the energy change.

Now just substitute the numbers given into the equation.

c = 2000 / 2 x 5
c = 2000/ 10
c = 200

Therefore we can conclude that the specific heat capacity of the block is 200 Jkg^-1°C^-1
3 0
3 years ago
A small boat coasts at constant speed under a bridge. A heavy sack of sand is dropped from the bridge onto the boat. The speed o
Tju [1.3M]

Answer:

d. decreases

Explanation:

The law of conservation of momentum tells us that the sum of momenta before the collision is equal to the sum of momenta after the collision. The bag has no momentum as it falls onto the boat because its velocity is zero in the horizontal direction. But after it hits the boat, it's momentum increases while the momentum of the system remains the same. That means a component of the system must decrease somewhere else. And that component is the velocity, not the mass, of the boat.

7 0
3 years ago
A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap
viktelen [127]

Answer:

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

V_r + V_L + V_c = V_{net}

now we will have

iR + L\frac{di}{dt} + \frac{q}{C} = 12 V

now we have

1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12

So we will have

q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}

at t = 0 we have

q = 0

0 = 3\times 10^{-6} + c_1  + c_2

also we know that

at t = 0 i = 0

0 = -4000 c_1 - 1000c_2

c_2 = -4c_1

c_1 = 1 \times 10^{-6}

c_2 = -4 \times 10^{-6}

so we have

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

5 0
3 years ago
What is the instantaneous velocity of the hummingbird at t=1s?
Liono4ka [1.6K]

The distance - time graph of the humming bird is missing, so i have attached it.

Answer:

Instantaneous velocity = 0.5 m/s

Explanation:

From the attached graph, at time t = 1 s, the corresponding distance is 0.5 m.

Instantaneous velocity is the velocity at that point.

Thus;

Instantaneous velocity = 0.5/1

Instantaneous velocity = 0.5 m/s

3 0
3 years ago
Consider a system of two particles: ball A with a mass m is moving to the right a speed 2v and ball B with a mass 3m is moving t
arlik [135]

Answer:

Explanation:

Answer:

Explanation:

Given that,

System of two particle

Ball A has mass

Ma = m

Ball A is moving to the right (positive x axis) with velocity of

Va = 2v •i

Ball B has a mass

Mb = 3m

Ball B is moving to left (negative x axis) with a velocity of

Vb = -v •i

Velocity of centre of mass Vcm?

Velocity of centre of mass can be calculated using

Vcm = 1/M ΣMi•Vi

Where M is sum of mass

M = M1 + M2 + M3 +...

Therefore,

Vcm=[1/(Ma + Mb)] × (Ma•Va +Mb•Vb

Rearranging for better understanding

Vcm = (Ma•Va + Mb•Vb) / ( Ma + Mb)

Vcm = (m•2v + 3m•-v) / (m + 3m)

Vcm = (2mv — 3mv) / 4m

Vcm = —mv / 4m

Vcm = —v / 4

Vcm = —¼V •i

3 0
3 years ago
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