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V125BC [204]
3 years ago
10

JUST TWO QUESTIONS PLEASEEEE HELPPPPPP IM SO LOSTTTT AND I HAVE TO SUBMIT ITTT

Physics
1 answer:
Scilla [17]3 years ago
7 0

Sry,i only the answer of that question no.1

(m)=1000kg

(u)=30m/s

(f)=5000n

(v)=0m/s

(a)=f/m

= 5000/1000

=5

Distance(s)=?

We know that,

V^2=u^2+2as

Or,0^2=30^2+2×5×s

Or,s=910m

So,the minium distance required to stop the car is 910m.

Hope,itwillhelpya!

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In midair an M = 145 kg bomb explodes into two pieces of m1 = 115 kg and another, respectively. Before the explosion, the bomb w
Daniel [21]

Answer:

v_2=-133.17m/s, the minus meaning west.

Explanation:

We know that linear momentum must be conserved, so it will be the same before (p_i) and after (p_f) the explosion. We will take the east direction as positive.

Before the explosion we have p_i=m_iv_i=Mv_i.

After the explosion we have pieces 1 and 2, so p_f=m_1v_1+m_2v_2.

These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.

Since we know momentum must be conserved we have:

Mv_i=m_1v_1+m_2v_2

Which means (since we want v_2 and M=m_1+m_2):

v_2=\frac{Mv_i-m_1v_1}{m_2}=\frac{Mv_i-m_1v_1}{M-m_1}

So for our values we have:

v_2=\frac{(145kg)(24m/s)-(115kg)(65m/s)}{(145kg-115kg)}=-133.17m/s

5 0
3 years ago
A. What is the RMS speed of Helium atoms when the temperature of the Helium gas is 343.0 K? (Possibly useful constants: the atom
kkurt [141]

Answer:

(a) 1462.38 m/s

(b) 2068.13 m/s

Explanation:

(a)

The Kinetic energy of the atom can be given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 10⁻²³ J/k

K.E = Kinetic Energy of atoms = 343 K

T = absolute temperature of atoms

The K.E is also given as:

K.E = (1/2)mv²

Comparing both equations:

(1/2)mv² = (3/2)KT

v² = 3KT/m

v = √[3KT/m]

where,

m = mass of Helium = (4 A.M.U)(1.66 X 10⁻²⁷ kg/ A.M.U) = 6.64 x 10⁻²⁷ kg

v = RMS Speed of Helium Atoms = ?

Therefore,

v = √[(3)(1.38 x 10⁻²³ J/K)(343 K)/(6.64 x 10⁻²⁷ kg)]

<u>v = 1462.38 m/s</u>

(b)

For double temperature:

T = 2 x 343 K = 686 K

all other data remains same:

v = √[(3)(1.38 x 10⁻²³ J/K)(686 K)/(6.64 x 10⁻²⁷ kg)]

<u>v = 2068.13 m/s</u>

8 0
3 years ago
What is the magnitude of the electric field in a region where the potential is given by the expression V = ax2 + b where a = −50
sweet [91]

Answer:

E = 1000 x

Explanation:

The electric potential and the electric field are related by the formula

        dV = - E . dx

Bold represents vectors.

The point represents the scalar product, in this case we calculate the electric field in the x-axis and the potential is also in this axis so the scalar product is reduced to the algebraic product

        E = dV /dx

Let's make the derivative

        E = - 2ax

Let's replace the values

        E = -2 (-500) x

        E = 1000 x

5 0
3 years ago
Read 2 more answers
Your toaster has a power cord with a resistance of 2.2x10^−2 Ω connected in series with a 9.7 Ω nichrome heating element. The po
kati45 [8]

Answer:

(A) 654.545 Kw

(B) 2.885\times 10^5kw

Explanation:

We have given resistance of the toaster R_1=2.2\times 10^{-2}ohm=0.022ohm

Resistance of nichrome heating element R_2=9.7ohm

Both the resistances are connected in series so same current will flow through the circuit

Potential difference across the toaster V = 120 volt

So current i=\frac{120}{0.022}=5454.5454A

(a) Power dissipated in toaster P=i^2R=5454.5454^2\times 0.022=654545.441W=654.545kw

(B) Power dissipated in heating element  P=i^2R=5454.5454^2\times 9.7=2.885\times 10^8W=2.885\times 10^5kw

5 0
4 years ago
Given the initial wavefunction Ψ (x, 0) = Axexp (-k x) withx&gt; 0 andk&gt; 0, and Ψ (x, 0) = 0 forx &lt;0, what value must A ta
madam [21]

Answer with explanation:

The Normalization Principle states that

\int_{-\infty }^{+\infty }f(x)dx=1

Given

f(x)=xe^{-kx}(x>0\\\\0(x

Thus solving the integral we get

\int_{0 }^{+\infty }A\cdot xe^{-kx}dx=1\\\\A\int_{0 }^{+\infty }\cdot xe^{-kx}dx=1

The integral shall be solved using chain rule initially and finally we shall apply the limits as shown below

I=\int xe^{-kx}dx\\\\x\int e^{-kx}dx-\int \frac{d(x)}{dx}\int e^{-kx}dx\\\\-\frac{xe^{-kx}}{k}-\int 1\cdot \frac{-e^{-kx}}{k}\\\\\therefore I=\frac{e^{-kx}}{k}-\frac{xe^{-kx}}{k}

Applying the limits and solving for A we get

I=\frac{1}{k}[\frac{1}{e^{kx}}-\frac{x}{e^{kx}}]_{0}^{+\infty }\\\\I=-\frac{1}{k}\\\\\therefore A=-k

3 0
3 years ago
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