What is the potential energy when a 100 kg object is raised 4.00 m straight up?

What is the speed of an object that travels 60 metres in 4 seconds

Zina [86]

Answer:

$s = \frac{d}{t} = \frac{60}{4} \\ \boxed{speed = 15m. {sec}^{ - 1} }$

4 0

2 years ago

Read 2 more answers

The total charge a battery can supply is rated in mA⋅h , the product of the current (in mA ) and the time (in h ) that the batte

natita [175]

Answer: 0.2 hours

Explanation: In order to solve this question we have to considerer that a recargeable battery can supply 1800 mA in one hour then we have to determine how long could this battery drive current through a long, thin wire of resistance 34 Ω .

Besides, this battery has a voltage of 12 V

so by using the Ohm law we also know that V=R*I,

Fron this we can obtain:

I= V/R= 12 V/ 34 Ω=0.35 A= 350 mA

then considering that this battery can supply 1800 mA in one hour we have this battery can supply 350 mA in x time in the form:

1hour------- 1800 mA

x hour--------350 mA

time= 350/1800= 0.2 hour

4 0

3 years ago

Do you have to keep citing the same source

emmasim [6.3K]

I don't think so as long as you make it apparent that the information comes the same source. So citing over and over again is unnecessary as long as it's clear that the information is from the same website or source. If you can't make it clear that they are from the same website source, it would a safe choice to continue to cite to avoid allegations of plagiarism.

4 0

2 years ago

Part A Determine the magnitude of the x component of F using scalar notation. Fx F x = nothing lb Request Answer Part B Determin

maksim [4K]

As we know that force F makes an angle of 60 degree with X axis

so the X component is given as

$cos60 = \frac{F_x}{F}$

now we have

$F_x = F cos60$

$F_x = 0.50 F$

Similarly we know that force F makes an angle of 45 degree with Y axis

so the X component is given as

$cos45 = \frac{F_y}{F}$

now we have

$F_y = F cos45$

$F_y = 0.707 F$

Now for the component along z axis we know that

$F_x^2 + F_y^2 + F_z^2 = F^2$

now plug in all components

$(0.707 F)^2 + (0.50 F)^2 + F_z^2 = F^2$

$0.5 F^2 + 0.25 F^2 + F_z^2 = F^2$

$F_z^2 = F^2(1 - 0.75)$

$F_z^2 = 0.25 F^2$

$F_z = 0.5 F$

5 0

3 years ago

Question 2

Delvig [45]

Answer:

Approximately $73\; {\rm N}$ , assuming that the acceleration of this ball is constant during the descent.

Explanation:

Assume that the acceleration of this ball, $a$ , is constant during the entire descent.

Let $x$ denote the displacement of this ball and let $t$ denote the duration of the descent. The SUVAT equation $x = (1/2)\, a\, t^{2}$ would apply.

Rearrange this equation to find an expression for the acceleration, $a$ , of this ball:

$\begin{aligned} a &= \frac{2\, x}{t^{2}}\end{aligned}$ .

Note that $x = 11\; {\rm m}$ and $t = 1.5\; {\rm s}$ in this question. Thus:

$\begin{aligned} a &= \frac{2\, x}{t^{2}} \\ &= \frac{2 \times 11\; {\rm m}}{(1.5\; {\rm s})^{2}} \\ &\approx 9.78\; {\rm m \cdot s^{-2}}\end{aligned}$ .

Let $m$ denote the mass of this ball. By Newton's Second Law of Motion, if the acceleration of this ball is $a$ , the net external force on this ball would be $m\, a$ .

Since $m = 7.5\; {\rm kg}$ and $a \approx 9.78\; {\rm m\cdot s^{-2}}$ , the net external force on this ball would be:

$\begin{aligned} (\text{net force}) &= m\, a \\ &\approx 7.5\; {\rm kg} \times 9.78\; {\rm m\cdot s^{-2}} \\ &\approx 73\; {\rm kg \cdot m \cdot s^{-2} \\ &= 73\; {\rm N} && (1\; {\rm N} = 1\; {\rm kg \cdot m\cdot s^{-2}}) \end{aligned}$ .

4 0

2 years ago