A metal object is four feet away from a magnet. If you move the object two feet closer toward the

6. Mass affects how fast an object falls.<br> True<br> False

dimulka [17.4K]

The statement: Mass affects how fast an object falls is true.

6 0

2 years ago

Suppose that in a lightning flash the potential difference between a cloud and the ground is 1.0*109 V and the quantity of charg

nata0808 [166]

Answer:

a) $U_{e} = 3 \times 10^{10}\,J$ , b) $v \approx 7745.967\,\frac{m}{s}$

Explanation:

a) The potential energy is:

$U_{e} = Q \cdot \Delta V$

$U_{e} = (30\,C)\cdot (1.0\times 10^{9}\,V)$

$U_{e} = 3 \times 10^{10}\,J$

b) Maximum final speed:

$U_{e} = \frac{1}{2}\cdot m \cdot v^{2}\\v = \sqrt{\frac{2\cdot U_{e}}{m} }$

The final speed is:

$v=\sqrt{\frac{2\cdot (3 \times 10^{10}\,J)}{1000\,kg} }$

$v \approx 7745.967\,\frac{m}{s}$

3 0

3 years ago

A cheetah can run at a maximum speed 97.8 km/h and a gazelle can run at a maximum speed of 78.2 km/h. If both animals are runnin

emmainna [20.7K]

(1)

Cheetah speed: $v_c = 97.8 km/h=27.2 m/s$

Its position at time t is given by

$S_c (t)= v_c t$ (1)

Gazelle speed: $v_g = 78.2 km/h=21.7 m/s$

the gazelle starts S0=96.8 m ahead, therefore its position at time t is given by

$S_g(t)=S_0 +v_g t$ (2)

The cheetah reaches the gazelle when $S_c=S_g$ . Therefore, equalizing (1) and (2) and solving for t, we find the time the cheetah needs to catch the gazelle:

$v_c t=S_0 + v_g t$

$(v_c -v_g t)=S_0$

$t=\frac{S_0}{v_c-v_t}=\frac{96.8 m}{27.2 m/s-21.7 m/s}=17.6 s$

(2) To solve the problem, we have to calculate the distance that the two animals can cover in t=7.5 s.

Cheetah: $S_c = v_c t =(27.2 m/s)(7.5 s)=204 m$

Gazelle: $S_g = v_g t =(21.7 m/s)(7.5 s)=162.8 m$

So, the gazelle should be ahead of the cheetah of at least

$d=S_c -S_g =204 m-162.8 m=41.2 m$

4 0

3 years ago

Read 2 more answers

A 3.0 kg block is pushed by a 14 N force. If µ = 0.6, will the block move?

Anna71 [15]

Answer:

The block will not move.

Explanation:

We'll begin by calculating the frictional force. This can be obtained as follow:

Coefficient of friction (µ) = 0.6

Mass of block (m) = 3 Kg

Acceleration due to gravity (g) = 10 m/s²

Normal reaction (R) = mg = 3 × 10 = 30 N

Frictional force (Fբ) =?

Fբ = µR

Fբ = 0.6 × 30

Fբ = 18 N

From the calculations made above, the frictional force of the block is 18 N. Since the frictional force (i.e 18 N) is bigger than the force applied (i.e 14 N), the block will not move.

4 0

2 years ago

A professor, with dumbbells in his hands and holding his arms out, is spinning on a turntable with an angular velocity. What hap

Olenka [21]

Answer:

<em>His angular velocity will increase.</em>

Explanation:

According to the conservation of rotational momentum, the initial angular momentum of a system must be equal to the final angular momentum of the system.

The angular momentum of a system = $I$ 'ω'

where

$I$ ' is the initial rotational inertia

ω' is the initial angular velocity

the rotational inertia = $mr'^{2}$

where m is the mass of the system

and r' is the initial radius of rotation

Note that the professor does not change his position about the axis of rotation, so we are working relative to the dumbbells.

we can see that with the mass of the dumbbells remaining constant, if we reduce the radius of rotation of the dumbbells to r, the rotational inertia will reduce to $I$ .

From

$I$ 'ω' = $I$ ω

since $I$ is now reduced, ω will be greater than ω'

therefore, the angular velocity increases.

5 0

3 years ago