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Irina18 [472]
2 years ago
7

Consider a box sitting in the back of a pickup. The pickup accelerates to the right, and because the bed of the pickup is sticky

, the box does not slide around the truck when this happens.What direction is the force acting on the box due to the truck?
Physics
1 answer:
andreyandreev [35.5K]2 years ago
8 0
The force would be coming from the right causing the box the lean/ slide to left, if it wasnt sticky.
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A basketball with mass of 0.8 kg is moving to the right with velocity 6 m/s and hits a volleyball with mass of 0.6 kg that stays
IceJOKER [234]

Answer:

26.67 m/s

Explanation:

From the law of conservation of linear momentum, the initial sum of momentum equals the final sum.

p=mv where p is momentum, m is the mass of object and v is the speed of the object

Initial momentum

The initial momentum will be that of basketball and volleyball, Since basketball is initially at rest, its initial velocity is zero

p_i= m_bv_b+m_vv_v=8*6+0.6*0=48 Kg.m/s

Final momentum

p_f= m_bv_b+m_vv_v=8*4+0.6*v_v=32+0.6v Kg.m/s\\32+0.6v_v=48\\0.6v=16\\v_v=16/0.6=26.66666667\approx 26.67 m/s

4 0
2 years ago
A plane initially traveling at 200 m/s due west experiences a 10 m/s head wind coming from the opposite direction. A). What will
Hitman42 [59]
Ok so it would be late and the relative velocity would be 190 m/s because 200 m/s - 10 m/s is 190 m/s. Hope this helps.
8 0
2 years ago
"As the Voyager spacecraft penetrated into the outer solar system, the illumination from the Sun declined. Relative to the situa
____ [38]

Answer:

\frac{I_{2}}{I_{1}} = 0.04

Explanation:

The intensity of a star noticed at a certain distance is inversely proportional to the square of distance. Then:

I_{1}\cdot r_{1}^{2} = I_{2}\cdot r_{2}^{2}

The intensity of the Sun in Jupiter relative to Earth is:

\frac{I_{2}}{I_{1}} = \frac{r_{1}^{2}}{r_{2}^{2}}

\frac{I_{2}}{I_{1}} = \left(\frac{1\,AU}{5.2\,AU} \right) ^{2}

\frac{I_{2}}{I_{1}} = 0.04

3 0
2 years ago
Steam undergoes an adiabatic expansion in a piston–cylinder assembly from 100 bar, 360°C to 1 bar, 160°C. What is work in kJ per
vfiekz [6]

Answer:

work is 130.5 kJ/kg

entropy change is 1.655 kJ/kg-k

maximum  theoretical work is 689.4 kJ/kg

Explanation:

piston cylinder assembly

100 bar, 360°C to 1 bar, 160°C

to find out

work  and amount of entropy  and magnitude

solution

first we calculate work i.e heat transfer - work =   specific internal energy @1 bar, 160°C  - specific internal energy @ 100 bar, 360°C    .................1

so first we get some value from steam table with the help of 100 bar @360°C and  1 bar @ 160°C

specific volume = 0.0233 m³/kg

specific enthalpy = 2961 kJ/kg

specific internal energy = 2728 kJ/kg

specific entropy = 6.004 kJ/kg-k

and respectively

specific volume = 1.9838 m³/kg

specific enthalpy = 2795.8 kJ/kg

specific internal energy = 2597.5 kJ/kg

specific entropy = 7.659 kJ/kg-k

now from equation 1 we know heat transfer q = 0

so - w =   specific internal energy @1 bar, 160°C  - specific internal energy @ 100 bar, 360°C

work = 2728 - 2597.5

work is 130.5 kJ/kg

and entropy change formula is i.e.

entropy change =  specific entropy ( 100 bar @360°C)  - specific entropy ( 1 bar @160°C )

put these value we get

entropy change =  7.659 - 6.004

entropy change is 1.655 kJ/kg-k

and we know maximum  theoretical work = isentropic work

from steam table we know specific internal energy is 2038.3 kJ/kg

maximum  theoretical work = specific internal energy - 2038.3

maximum  theoretical work = 2728 - 2038.3

maximum  theoretical work is 689.4 kJ/kg

3 0
2 years ago
If the envelope and gondola have a total mass of 4300 kg, what is the maximum cargo load when the blimp flies at a sea-level loc
geniusboy [140]

Complete question:

The classic Goodyear blimp is essentially a helium balloon— a big one, containing 5700 m³ of helium. If the envelope and gondola have a total mass of 4300 kg, what is the maximum cargo load when the blimp flies at a sea-level location? Assume an air temperature of 20°C.

Answer:

52.4 kN

Explanation:

The helium at 20°C has a density of 0.183 kg/m³, and the cargo load is the weight of the system, which consists of the envelope, the gondola, and the helium.

The helium mass is the volume multiplied by the density, thus:

mHe = 5700 * 0.183 = 1043.1 kg

The total mass is then 5343.1 kg. The weight is the mass multiplied by the gravity acceleration (9.8 m/s²), so:

W = 5343.1*9.8

W = 53362.38 N

W = 52.4 kN

5 0
2 years ago
Read 2 more answers
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