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4 years ago

A_____ is the amount of heat needed to raise the temperature of 1 kilogram of water 1 degree celsius.

Physics

2 answers:

matrenka [14]4 years ago

7 0

Answer:

Calorie is the amount of heat needed to raise the temperature of 1 kilogram of water 1 degree Celsius.

Explanation:

Rom4ik [11]4 years ago

5 0

<span>A_____ is the amount of heat needed to raise the temperature of 1 kilogram of water 1 degree celsius.

Specific Heat Capacity of Water.
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An initially stationary 1.3 kg object accelerates horizontally and uniformly to a speed of 9.4 m/s in 3.0 s. (a) In that 3.0 s i

ehidna [41]

Answer: 57.434 Nm

Explanation: mass of object = 1.3kg, initial speed = 0m/s ( since it was at rest), final speed = 9.4 m/s.

Using the work energy theorem,

The work done on the object equals it kinetic energy.

Work done = mv²/2

Work done = 1.3 ×( 9.4 - 0)²/2

Work done = 1.3 × 9.4² / 2

Work done = 114.868 / 2 = 57.434 Nm

7 0

3 years ago

A physics student swings a tennis ball connected to a rope in a vertical circle with a constant speed of 6.29 m/s. The ball has

Alex777 [14]

Answer:

r = 0.5 m

Explanation:

First we find the angular speed of the ball by using its period:

ω = θ/t

For the time period:

ω = angular speed = ?

θ = angular displacement = 2π rad

t = time period = 0.5 s

Therefore,

ω = 2π rad/0.5 s

ω = 12.56 rad/s

Now, for the radius:

v = rω

r = v/ω

where,

v = linear speed = 6.29 m/s

r = radius = ?

r = (6.29 m/s)/(12.56 rad/s)

8 0

3 years ago

Collar P slides outward at a constant relative speed along rod AB, which rotates counterclockwise with a constant angular veloci

diamong [38]

Answer:

a= 23.65 ft/s²

Explanation:

given

r= 14.34m

ω=3.65rad/s

Ф=Ф₀ + ωt

t = Ф - Ф₀/ω

= (98-0)× $\frac{\pi}{180}$ /3.65

98°= 1.71042 rad

1.7104/3.65

t= 0.47 s

r₁(not given)

assuming r₁ =20 in

r₁ = r₀ + ut(uniform motion)

u = r₁ - r₀/t

r₀ = 14.34 in= 1.195 ft

r₁ = 20 in = 1.67 ft

= (1.667 - 1.195)/0.47

0.472/0.47

u= 1.00ft/s

acceleration at collar p

a=rω²

= 1.67 × 3.65²

a = 22.25ft/s²

acceleration of collar p related to the rod = 0

coriolis acceleration = 2ωu

= 2× 3.65×1 = 7.3 ft/s²

acceleration of collar p

= 22.5j + 0 + 7.3i

√(22.5² + 7.3²)

the magnitude of the acceleration of the collar P just as it reaches B in ft/s²

a= 23.65 ft/s²

4 0

3 years ago

Richard Julius once made a model plane that could travel a max speed of 110 m/s. Suppose the plane was held in a circular path b

hjlf

Answer:

85.8 m/s

Explanation:

We know that the length of the circular path, L the plane travels is

L = rθ where r = radius of path and θ = angle covered

Now,its speed , v = dL/dt = drθ/dt = rdθ/dt + θdr/dt

where dθ/dt = ω = angular speed = v'/r where v' = maximum speed of plane and r = radius of circular path

Now, from θ = θ₀ + ωt where θ₀ = 0 rad, ω = angular speed and t = time,

θ = θ₀ + ωt = 0 + ωt = ωt

So, v = rdθ/dt + θdr/dt

v = rω + ωtdr/dt

v = (r + tdr/dt)ω

v = (r + tdr/dt)v'/r

v = v' + tv'/r(dr/dt)

v = v'[1 + t(dr/dt)/r]

Given that v' = 110 m/s, t = 33.0s, r = 120 m and dr/dt = rate at which line is shortened = -0.80 m/s (negative since it is decreasing)

So, v = 110 m/s[1 + 33.0 s(-0.80 m/s)/120 m]

v = 110 m/s[1 + 11.0 s(-0.80 m/s)/40 m]

v = 110 m/s[1 + 11.0 s(-0.02/s)]

v = 110 m/s[1 - 0.22]

v = 110 m/s(0.78)

v = 85.8 m/s

8 0

3 years ago

The surface tension of water was determuned in a laboratory by using the drop weight method. 100 drops were released from a bure

lana [24]

Answer:

The surface tension of the water is 6.278×10⁻² N/m

error = 13.65%

Explanation:

The surface tension of water is given by

$\gamma = \frac{F}{L}$

Where F is the force acting on water and L is the length over which is force is acted.

We are given the mass of 100 droplets of water

M = 3.78 g

n = 100

The mass of 1 droplet is given by

$m = \frac{M}{n} \\\\m = \frac{3.78}{100}\\\\m = 0.0378 \: g \\\\m = 3.780\times10^{-5} \: kg$

The force acting on a single droplet of water is given by

$F = m \cdot g$

Where m is the mass of water droplet and g is the acceleration due to gravity

$F = 3.780\times10^{-5} \cdot 9.81$

$F = 3.708\times10^{-4} \: N$

The circumferential length of the droplet is given by

$L = \pi \cdot d$

Where d is the diameter

$L = \pi \cdot 1.88\times10^{-3}\\\\L = 5.906 \times10^{-3} \: m$

Now we can find out the required surface tension of the water

$\gamma = \frac{3.708\times10^{-4} }{5.906 \times10^{-3}} \\\\\gamma = 0.06278\: N/m\\\\\gamma = 6.278 \times10^{-2} \: N/m\\\\$

Therefore, the surface tension of the water is 6.278×10⁻² N/m

The tabulated value of the surface tension of water at 20 °C is given by

$\gamma_t = 0.0727 \: N/m$

The percentage error between tabulated and calculated surface tension is given by

$error = \frac{\gamma_t - \gamma }{\gamma_t}$

$error = \frac{ 0.0727 - 0.06278}{0.0727} \times 100\%$

$$ error = 13.65 \%$

7 0

3 years ago