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3 years ago

Which tools would Sara use to find and irregularly shaped objects mass in volume?

1 answer:

8 0

To find the mass and volume of irregularly shaped object. One can use the balance to find the weight, and to find the volume of the object water should be poured in the beaker and note the level of water.

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An elevator provides 21 000 w of power during a 12 s ride. how much work does the elevator do?

Mice21 [21]

Answer:

252000 J

Explanation:

W = Pt

W = (21000 w)(12 s)

W =252000 J

4 0

3 years ago

Consider the sections of two circuits illustrated above. Select True or False for all statements. After connecting c and d to a

Basile [38]

Answer:

Follows are the solution to the given question:

Explanation:

In this question the missing file of the circuit is not be which is defined in attached file please find it.

In Option 1, this statement is true because the current is on $R_3 \ and \ R_4$ , that is the same.

In option 2, this statement is false because $Rcd=R_3+R_4$ therefore it implies that Rcd is always larger then $R_3$ .

In option 3, this statement is true because the voltage of $R_1 \ and \ R_2$ is always equal.

In option 4, this statement is true because $Rab= \frac{1}{(\frac{1}{R1}+\frac{1}{R2})}=\frac{R1R2}{(R1+R2)}. \frac{R2}{(R1+R2)}$ is always smaller then 1 therefore, $R_1 \times (\frac{R2}{(R1+R2)})$ is always equal to R1.

5 0

3 years ago

The driver accelerates a 330.0 kg snowmobile, which results in a force being exerted that speeds up the snowmobile from 6.00 m/s

just olya [345]

Answer:

(a) 5610 kgm/s

(b) 5610 Ns.

(c) 78. 64 N

Explanation:

a. Change in momentum: This can be defined as the product of the mass of a body to its change in velocity. The S.I unit of change in momentum is kgm/s.

Mathematically, change in momentum is expressed as

ΔM = mΔv......................... Equation 1

Where ΔM = change in momentum, m = mass of snowmobile, Δv = change in velocity.

Given: m = 330 kg, Δv = v₂-v₁ = 23-6 = 17 m/s.

Note: v₁ and v₂ are the initial and the final velocity of the snowmobile.

ΔM = 330(17)

ΔM = 5610 kgm/s.

(b) Impulse: This can be defined as the product and force and time. The S.I unit of impulse is Ns.

Note: From Newton's second law of motion, impulse is equal to change in momentum.

Therefore,

I = ΔM................ Equation 2

Where I = impulse of the force.

Since ΔM = 5610 kgm/s.

Therefore

I = 5610 Ns.

Thus the impulse = 5610 Ns.

(c) Force: This can be defined as the product of the mass of a body and its acceleration. The S.I unit of force is Newton (N).

F = ma ................................. Equation 3

F = force, m = mass of the body, a = acceleration

But,

a = ( v₂-v₁)/t

Where v₂ = 23.0 m/s, v₁ = 6.0 m/s t = 60.0 s.

a = (23-6)/60

a = 0.283 m/s².

Substituting the value a and m into equation 3

F = 330(0.2383)

F = 78.639 N.

F ≈ 78. 64 N

8 0

3 years ago

Be sure to answer all parts. By what factor does the fraction of collisions with energy equal to or greater than activation ener

qwelly [4]

Answer:

Factor = 8.77

Explanation:

Fraction of collision with energy greater than or equal to the activation energy can be given by the formula:

$f = \exp(\frac{-E_{a} }{RT} )$

$E_{a} = Activation Energy\\E_{a} = 100 kJ /mol\\E_{a} = 10^{5} J /mol\\$

R = 8.314 J/mol.K

When Temperature = 34⁰C = 34 + 273 = 307 K

$f_{1} = \exp(\frac{-10^{5} }{8.314 * 307} )\\f_{1} = \exp(\frac{-10^{5} }{2552.398} )\\f_{1} = \exp(-39.18)\\f_{1} = 964.59 * 10^{-20}$

When Temperature = 52⁰C = 52 + 273 = 325 K

$f_{2} = \exp(\frac{-10^{5} }{8.314 * 325} )\\f_{2}= \exp(\frac{-10^{5} }{2702.05} )\\f_{2}= \exp(-37.009)\\f_{2} = 8456.6 * 10^{-20}$

$\frac{f_{2}}{f_{1}}= \frac{8456.6 * 10^{-20} }{964.59 * 10^{-20} }$

Factor = 8.77

8 0

3 years ago

A waiter is carrying a tray above his head and walking at a constant velocity. If he applies a force of 5.0 newtons on the tray

k0ka [10]

W = f × x = 5 × 10 = 50

4 0

3 years ago

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