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OLEGan [10]
2 years ago
10

A 0.05kg dart is thrown at and sticks into a 0.4 kg block hanging on a string. After the collision the block and dart swing in a

circular arc rising 0.1m vertically. What must the speed have been just after the collision
Physics
1 answer:
zloy xaker [14]2 years ago
6 0

Answer:

v = 1.4  m /s

Explanation:

We shall apply law of conservation of mechanical energy

The kinetic energy of dart and block   is converted into potential energy of both dart and block .

1 /2 (m+M) v² = ( m +M) gH

.5  x v² =  9.8 x .1

=  v² = 1.96

v = 1.4

v = 1.4  m /s

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3 0
2 years ago
when do you use cos and sin in situations like these? is horizontal always cos and vertical always sin?
Andreas93 [3]

Answer:

yes

Explanation:

this is simple

the horizontal line is adjacent

the vertical line is opposite

recall that cos x=adj/hyp

adj=hyp(cos x)

while opp=hyp(sin x)

8 0
3 years ago
the presence of which magnetic feature best explains why a magnet can act a distance on other magnets or on objects containing c
katen-ka-za [31]

Magnetic fields

Explanation:

The presence of magnetic fields  best explains why a magnet can act a distance on other magnets or on objects containing certain metals.

  • Magnetic fields are lines of forces around a bar magnet.
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8 0
3 years ago
A piston-cylinder device contains Helium gas initially at 150 kPa, 20 o C, and 0.5m 3 . The helium is now compressed in a polytr
Molodets [167]

Answer:

Explanation:

Given

P_1=150 kPa

T_1=20^{\circ}C

V_1=0.5 m^3

T_2=140^{\circ}C

P_2=400 kPa

R for Helium R=2.076

c_v=3.115 kJ/kg-K

mass of gas m=\frac{P_1V_1}{RT_1}

m=\frac{150\times 0.5}{2.076\times 293}

m=0.123 kg

Similarly V_2 can be found

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

V_2=0.264 m^3

Work done W=\int_{V_1}^{V_2}PdV

W=\frac{P_2V_2-P_1V_1}{n-1}

W=\frac{mR(T_2_T_1)}{n-1}

Since it is a polytropic Process

therefore PV^n=c

P_1V_1^n=P_2V_2^n

(\frac{V_1}{V_2})^n=\frac{P_2}{P_1}

(\frac{0.5}{0.264})^n=\frac{400}{150}

n=\frac{\ln 2.66}{\ln 1.893}

n=1.533

W=\frac{0.123\times 2.076(140-20)}{1.533-1}

W=57.48 kJ    

From Energy balance

E_{in}-E_{out}=\Delta E_{system}

Neglecting kinetic and Potential Energy change

Q_{in}+W_{in}=change\ in\ Internal\ Energy

Change in Internal Energy \Delta U=u_2-u_1

\Delta U=mc_v(T_2-T_1)

\Delta U=0.123\times 3.115(140-20)

\Delta U=45.977 kJ

Q_{in}+57.48=45.977

Q_{in}=-11.50 kJ  

i.e. Heat is being removed

3 0
3 years ago
A test charge is placed at a distance of 2.5 × 10-2 meters from a charge of 6.4 × 10-5 coulombs. What is the electric field at t
Novosadov [1.4K]
Using the formula: E = kQ / d² where E is the electric field, Q is the test charge in coulomb, and d is the distance. 

E = kQ / d²

k = 9 x 10^9 N-m²/C²
Q = 6.4 x 10^-5 C
d = 2.5 x 10^-2 m

Substituting the given values to the equation, we have:
E = (9 x 10^9)(6.4 x 10^-5) / (2.5 x 10^-2) ²

Electric field at the test charge is 921600000 N/C
8 0
3 years ago
Read 2 more answers
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