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kolbaska11 [484]
3 years ago
15

In which decade was Sputnik I launched? A. 1930-1939 B. 1950-1959 C. 1960-1969 D. 1980-1989

Physics
1 answer:
Olenka [21]3 years ago
6 0
B) 1950-1959
Sputnik I was launched on October 4, 1957
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A good baseball pitcher can throw a baseball toward home plate at 87 mi/h with a spin of 1710 rev/min. How many revolutions does
otez555 [7]

Explanation:

First, convert 87 mi/h to ft/min.

87 mi/h × (5280 ft/mi) × (1 h / 60 min) = 7656 ft/min

The time to reach the home plate is:

t = 60 ft / 7656 ft/min

t = 0.00784 min

The number of revolutions made in that time is:

n = 1710 rev/min × 0.00784 min

n = 13.4 rev

Rounding to 2 significant figures, the ball makes 13 revolutions.

4 0
3 years ago
A small ball of mass m is aligned above a larger ball of mass M = 0.63kg (with a slight separation) and the two are dropped simu
natta225 [31]
 <span>a) M is the big one 
m is the little one 

v is the speed of each of them when they impact 

V = speed of mii after collision 

Conservation of momentum 

Mv – mv = mV 

Conservation of energy 

1/2Mv^2 + 1/2mv^2 = 1/2 mV^2 

This pair simplify to give 

M = 3m 

V = 2v 

So m = 0.21kg 

and h = 4 . 2.7 = 10.8 m</span>Source(s):<span>Old teacher</span>
8 0
3 years ago
A 16.0 Ω, 13.0 Ω, and 7.00 Ω resistor are connected in parallel to an emf source. A current of 6.00 A is in the 13.0 Ω resistor.
Darya [45]

Answer:3.54ohms

Explanation: connection in parallel

1/Rt= 1/R1+1/R2+1/R3

1/Rt= 1/16+1/13+1/7

1/Rt= 91+112+208/1456

1/Rt= 411/1456

411Rt= 1456

Rt= 1456/411

Rt= 3.54ohms

3 0
4 years ago
Read 2 more answers
Suppose that the coefficient of kinetic friction between Zak's feet and the floor, while wearing socks, is 0.250. Knowing this,
insens350 [35]

Answer:

The distance travel before stopping is 1.84 m

Explanation:

Given :

coefficient of kinetic friction \mu_{k} = 0.250

Zak's speed v = 3 \frac{m}{s}

Gravitational acceleration g = 9.8 \frac{m}{s^{2} }

Work done by frictional force is given by,

  W = \Delta K

 \mu _{k} mg d = \frac{1}{2} m v^{2}

  d = \frac{v^{2}  }{2 g \mu _{k} }

  d = \frac{9}{2 \times 9.8 \times 0.250}

  d = 1.84 m

Therefore, the distance travel before stopping is 1.84 m

3 0
3 years ago
HELP ME PLEASEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE A student uses a spring scale attached to a textbook to compa
aleksandrvk [35]

Answer:

1,860

Explanation:

4 0
3 years ago
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