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3 years ago

PLS I NEED HELP ASAAP

Physics

1 answer:

Lerok [7]3 years ago

5 0

Answer:

Option A

Explanation:

The Equation represents the displacement of the object which is represented by x

$x=t-t^2$

so, $x_0$ means when time is zero so we replace t with zero in the equation,

$x_0=(0)-(0)^2\\x_0=0$

now for v which is velocity we need to differentiate the function as the formula for velocity is rate of change of displacement over time so we derivate the equation once and get,

$v=1-2t\\$

now for $v_0$ we insert t = 0 and get

$v_0=1-2(0)\\v_0=1$

now for a which is acceleration the formula of acceleration is rate of change of velocity over time, so we differentiate the the equation of v(velocity) once or the equation of x(displacement) twice so now we get,

$a=-2$

so Option A is your answer.

Remember derivative of a constant is always zero because a constant value has no rate of change has its a constant hence the derivative is 0

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A weightlifter lifts a 250-kg mass 0.5 meters above his head, how much PEg does the mass have (Note: g=9.8 m/s2)? Round your ans

morpeh [17]

Answer:

1225 J

Explanation:

The Gravitational potential energy (PEG) gained by a mass lifted above the ground is given by

$PE=mgh$

where

m is the mass

g = 9.8 m/s^2 is the acceleration due to gravity

h is the height at which the object has been lifted

In this problem, we have

m = 250 kg

h = 0.5 m

So, the PE of the object is

$PE=(250 kg)(9.8 m/s^2)(0.5 m)=1225 J$

6 0

3 years ago

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Reflected waves change their wavelength by ______ when reflected.

natita [175]

Answer:

i think its c

Explanation:

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4 0

2 years ago

A plastic rod that has been charged to − 15 nC touches a metal sphere. Afterward, the rod's charge is − 5.0 nC.

Natali5045456 [20]

Answer:

B) electrons transferred from sphere to rod.

(2) 1.248 x 10¹¹ electrons were transferred

Explanation:

Given;

initial charge on the plastic rod, q₁ = 15nC

final charge on the plastic rod, q₂ = - 5nC

let the charge acquired by the plastic rod = q

q + 15nC = -5nC

q = -5nC - 15nC

q = -20 nC

Thus, the plastic rod acquired excess negative charge from the metal sphere.

Hence, electrons transferred from sphere to rod

B) electrons transferred from sphere to rod.

2) How many charged particles were transferred?

1.602 x 10⁻¹⁹ C = 1 electron

20 x 10⁻⁹ C = ?

= 1.248 x 10¹¹ electrons

Thus,1.248 x 10¹¹ electrons were transferred

7 0

3 years ago

The radius of a planet is 2400 km, and the acceleration due to gravity at its surface is 3.6 m/s2.

kiruha [24]

Answer:

$3.1\cdot10^{23}\:\mathrm{kg}$

Explanation:

We can use Newton's Universal Law of Gravitation to solve this problem:

$g_P=G\frac{m}{r^2}$ ., where $g_P$ is acceleration due to gravity at the planet's surface, $G$ is gravitational constant $6.67\cdot 10^{-11}$ , $m$ is the mass of the planet, and $r$ is the radius of the planet.

Since acceleration due to gravity is given as $m/s^2$ , our radius should be meters. Therefore, convert $2400$ kilometers to meters:

$2400\:\mathrm{km}=2,400,000\:\mathrm{m}$ .

Now plugging in our values, we get:

$3.6=6.67\cdot10^{-11}\frac{m}{(2,400,000)^2}$ ,

Solving for $m$ :

$m=\frac{2,400,000^2\cdot3.6}{6.67\cdot 10^{-11}},\\m=\fbox{$3.1\cdot10^{23}\:\mathrm{kg}$}$ .

6 0

2 years ago

A rugby player sits on a scrum machine that weighs 200 Newtons. Given that the coefficient of static friction is 0.67, the coeff

Trava [24]

a. 850 N is the minimum force needed to get the machine/player system moving, which means this is the maximum magnitude of static friction between the system and the surface they stand on.

By Newton's second law, at the moment right before the system starts to move,

• net horizontal force

∑ F[h] = F[push] - F[s. friction] = 0

• net vertical force

∑ F[v] = F[normal] - F[weight] = 0

and we have

F[s. friction] = µ[s] F[normal]

It follows that

F[weight] = F[normal] = (850 N) / (0.67) = 1268.66 N

where F[weight] is the combined weight of the player and machine. We're given the machine's weight is 200 N, so the player weighs 1068.66 N and hence has a mass of

(1068.66 N) / g ≈ 110 kg

b. To keep the system moving at a constant speed, the second-law equations from part (a) change only slightly to

∑ F[h] = F[push] - F[k. friction] = 0

∑ F[v] = F[normal] - F[weight] = 0

so that

F[k. friction] = µ[k] F[normal] = 0.56 (1268.66 N) = 710.45 N

and so the minimum force needed to keep the system moving is

F[push] = 710.45 N ≈ 710 N

4 0

2 years ago

PLS I NEED HELP ASAAP​

PLS I NEED HELP ASAAP