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3 years ago

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An object has a mass of 15 kg. on planet x, the object weighs 126 n. the force of gravity on planet x is _____. 0.12 m/s2 8.4 m/

s2 1875 m/s2

2 answers:

alexandr402 [8]3 years ago

4 0

<span>8.4 m/s^<span>2
Is correct!</span></span>

Lana71 [14]3 years ago

4 0

Answer:

B. 8.4 m/s^2

Explanation:

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nasty-shy [4]

Answer:

all qn 1,2,3 have same answer ,. Yes,. hope it helps

3 0

3 years ago

The velocity of the transverse waves produced by an earthquake is 5.05 km/s, while that of the longitudinal waves is 8.585 km/s.

sattari [20]

Answer:

$d=691.71km$

Explanation:

The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:

$t=\frac{d}{v_t}-\frac{d}{v_l}$

Here d is the distance at which the earthquake take place and $v_t, v_l$ is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:

$t=d(\frac{1}{v_t}-\frac{1}{v_l})\\d=\frac{t}{\frac{1}{v_t}-\frac{1}{v_l}}\\d=\frac{56.4s}{\frac{1}{5.05\frac{km}{s}}-\frac{1}{8.585\frac{km}{s}}}\\d=691.71km$

8 0

3 years ago

A magnetic B field of strength 0.9 T is perpendicular to a loop with an area of 3 m2. If the area of the loop is reduced to zero

Alexeev081 [22]

Answer:

The magnitude of induced emf is 5.4 V

Explanation:

Given:

Magnetic field $B = 0.9$ T

Area of loop $\Delta A = 3$ $m^{2}$

Time take to reduce loop to zero $\Delta t = 0.5$ sec

To find induced emf we use faraday's law,

Induced emf is given by,

$\epsilon = -\frac{\Delta \phi}{\Delta t}$

Here minus sign shows lenz law, for finding magnitude of emf we ignore it.

Where $\phi = B\Delta A$

Put the value of flux and find induced emf,

$\epsilon = \frac{B\Delta A}{\Delta t}$

$\epsilon = \frac{0.9 \times 3}{0.5}$

$\epsilon = 5.4$ V

Therefore, the magnitude of induced emf is 5.4 V

3 0

4 years ago

If the mass is displaced 0.270 m from equilibrium and released, what is its speed when it is 0.130 m from equilibrium?

puteri [66]

Answer:

The speed is 1.52 m.

Explanation:

Given that,

Displacement =0.270 m

Distance = 0.130 m

Suppose a 0.321-kg mass is attached to a spring with a force constant of 13.3 N/m.

We need to calculate the angular velocity

Using formula of angular velocity

$\omega=\sqrt{\dfrac{k}{m}}$

Put the value into the formula

$\omega=\sqrt{\dfrac{13.3}{0.321}}$

$\omega=6.43\ rad/s$

We need to calculate the velocity

Using formula of velocity

$v=\omega\sqrt{A^2-x^2}$

Put the value into the formula

$v=6.43\times\sqrt{0.270^2-0.130^2}$

$v=1.52\ m/s$

Hence, The speed is 1.52 m.

6 0

3 years ago

How protons does Cl-38 have with the atomic number being 38

denis-greek [22]

Answer:

Explanation:

17

6 0

3 years ago