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3 years ago

A baby elephant weighs 250 pounds on Earth. How much would the elephant weigh on Saturn?

1 answer:

8 0

Well since staurns gravity is 10.44 multiply the elephants weight times the gravity
250x10.44 you will get 2610
hope i helped add me
Dosvedanya :)

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The average coefficient of linear expansion of copper is 1.7 10-5 (°c)−1. the statue of liberty is 93 m tall on a summer morning

sammy [17]

Let the rise in temperature be $5^0C$

The expansion in length due to change in temperature is given by the expression lαΔt , where l is the length, α is the coefficient of linear expansion, Δt is the change in temperature.

Here l = 93 m, α = $1.7*10^{-5}$ $^0C^{-1}$ , and Δt = $5^0C$

So expansion in length = 93* $1.7*10^{-5}$ *5 = 0.007905 m = $0.79*10^{-3}m$

So order of magnitude in change in length = -3

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3 years ago

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A series circuit consists of a 100-ω resistor, a 10.0-μf capacitor, and a 0.350-h inductor. the circuit is connected to a 120-v

Tpy6a [65]

Current will be $I=\dfrac{V_{rms}}{Z}=\dfrac{V_{rms}}{\sqrt{ R^{2}+(X_{C}-X_{L})^{2}}}\\where~X_{c}=\dfrac{1}{j.\omega .C}~and~X_{L}=j.\omega.L~where~\omega=2.\pi f~and~f=60Hz$
now just pluf in the values and Voila..

7 0

3 years ago

The Young’s modulus of nickel is Y = 2 × 1011 N/m2 . Its molar mass is Mmolar = 0.059 kg and its density is rho = 8900 kg/m3 . G

Charra [1.4K]

Answer:

Atomic Size and Mass:

convert given density to kg/m^3 = 8900kg/m^3 2) convert to moles/m^3 (kg/m^3 * mol/kg) = 150847 mol/m^3 (not rounding in my actual calculations) 3) convert to atoms/m^3 (6.022^23 atoms/mol) = 9.084e28 atoms/m^3 4) take the cube root to get the number of atoms per meter, = 4495309334 atoms/m 5) take the reciprocal to get the diameter of an atom, = 2.2245e-10 m/atom 6) find the mass of one atom (kg/mol * mol/atoms) = 9.7974e-26 kg/atom Young's Modulus: Y=(F/A)/(dL/L) 1) F=mg = (45kg)(9.8N/kg) = 441 N 2) A = (0.0018m)^2 = 3.5344e-6 m^2 3) dL = 0.0016m 4) L = 2.44m 5) Y = 1.834e11 N/m^2 Interatomic Spring Stiffness: Ks,i = dY 1) From above, diameter of one atom = 2.2245e-10 m 2) From above, Y = 1.834e11 N/m^2 3) Ks,i = 40.799 N/m (not rounding in my actual calculations) Speed of Sound: v = ωd 1) ω = √(Ks,i / m,a) 2) From above, Ks,i = 40.799 N/m 3) From above, m,a = 9.7974e-26 kg 4) ω=2.0406e13 N/m*kg 5) From above, d=2.2245e-10 m 6) v=ωd = 4539 m/s (not rounding in actual calculations) Time Elapsed: 1) length sound traveled = L+dL = 2.44166 m 2) From above, speed of sound = 4539 m/s 3) T = (L+dL)/v = 0.000537505 s

7 0

3 years ago

A 6.0-kilogram block, sliding to the east across a horizontal, frictionless surface with a momentum of 30.0 kilogram · meters pe

Lina20 [59]

The final speed of the block after the collision with the obstacle is $\boxed{3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}$ .

Further Explanation:

Given:

The mass of the block is $6.0\,{\text{kg}}$ .

The initial momentum of the block is $30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ .

The impulse imparted by the obstacle is $10\,{\text{N}} \cdot {\text{s}}$ .

Concept:

The block is sliding towards east and the impulse imparted by the obstacle is towards the obstacle is towards west on the block. It means that the impulse exerted by the obstacle will reduce the momentum of the block.

According to the impulse momentum theorem, the rate of change of momentum of the body is equal to the impulse imparted to the body.

The expression for the impulse momentum theorem is.

${p_f} - p{ & _i} = I$ …… (1)

Substitute $30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ for ${p_i}$ and $- 10\,{\text{N}} \cdot {\text{s}}$ for $I$ in equation (1).

$\begin{aligned}{p_f} &= - 10\,{\text{N}} \cdot {\text{s}} + 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\&= 20\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$

The final momentum of the block can be expressed as:

${p_f} = m{v_f}$ …… (2)

Substitute $20\text{kg}\;\text{m/s}$ for ${p_f}$ and $6.0\,{\text{kg}}$ for $m$ in equation (2).

$\begin{aligned}20 &= 6 \times {v_f} \\ {v_f}&= \frac{{20}}{6}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\&= 3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} \\ \end{aligned}$

Thus, the final speed of the block after the collision with the obstacle is $\boxed{3.33\;\text{m/s}}$ .

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Answer Details:

Grade: High School

Chapter: Impulse-momentum theorem

Subject: Physics

Keywords: Impulse, imparted, obstacle, speed, momentum, the obstacle, impulse-momentum theorem, frictionless surface, speed of block after collision.

5 0

3 years ago

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A 0.10-kg ball, traveling horizontally at 25 m/s, strikes a wall and rebounds at 19 m/s. What is the 7) magnitude of the change

IRISSAK [1]

Answer:

Change in momentum will be -4.4 kgm/sec

So option (A) is correct option

Explanation:

Mass of the ball is given m = 0.10 kg

Initial velocity of ball $v_1=25m/sec$