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3 years ago

7

Who were we in the space/arms race with? In the movie Hidden figures

1 answer:

daser333 [38]3 years ago

4 0

Explanation:

The U.S. launched its first man into space in May 1961.

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2. This diagram represents a top-down view of an experiment on a table. The 250 g and 100 g masses are falling and are pulling t

Rasek [7]

Answer:

According to the data given in the question, experiment on table two pulling and falling masses are arranged in the fig. 250 g is pulling right side and 100 g pulling down. The gravitational force is common to both the masses, so we cannot say that the block moves towards heavier mass, also the block does not move towards the lighter mass.

Obviously, the effect of heavier mass of 250 g is more on the block, so the block moves towards right bottom corner. i.e., diagonally between two masses

please find the attachment.

4 0

3 years ago

Read 2 more answers

A 1090 kg car moving +11.0 m/s

stiks02 [169]

The mass of the second car is 1434.21 kg

<u>Explanation:</u>

Using law of conservation of momentum,

$m_{1} u_{1}+m_{2} u_{2}=\left(m_{1}+m_{2}\right) v$

Given:

$m_{1}$ = 1090 kg

$u_{1}$ = 11 m/s

$u_{2}$ = 0

v = 4.75 m/s

We need to find $m_{2}$

When substituting the given values in the above equation, we get

$(1090 \times 11)+\left(m_{2} \times 0\right)=\left(1090+m_{2}\right) 4.75$

$11990=5177.5+4.75 m_{2}$

$4.75 m_{2}=11990-5177.5$

$4.75 m_{2}=6812.5$

$m_{2}=\frac{6812.5}{4.75}=1434.21 \mathrm{kg}$

6 0

3 years ago

I don't get the flat line on a graph. on a distance/speed or motion graph, they say it is not moving. that is what I don't get i

timofeeve [1]

A flat line means the the speed is the same . Its moving at the same pace.

4 0

4 years ago

An 8.00 g bullet is fired into a 250 g block that is initially at rest at the edge of a table of 1.00 m height. The bullet remai

Vsevolod [243]

<span>79.75m/s .................................</span>

5 0

3 years ago

A 30-cm-diameter, 1.2 kg solid turntable rotates on a 1.2-cm-diameter, 450 g shaft at a constant 33 rpm. When you hit the stop s

skelet666 [1.2K]

Answer:

frictional force = 0.52 N

Explanation:

diameter of turn table (D1) = 30 cm = 0.3 m

mass of turn table (M1) = 1.2 kg

diameter of shaft (D2) = 1.2 cm = 0.012 m

mass of shaft (M2) = 450 g = 0.45 kg

time (t) = 15 seconds

acceleration due to gravity (g) = 9.8 m/s^{2}

radius of turn table (R1) = 0.3 / 2 = 0.15 m

radius of shaft (R2) = 0.012 / 2 = 0.006 m

total moment of inertia (I) = moment of inertia of turn table + moment of inertia of shaft

I = 0.5(M1)(R1)^{2} + O.5 (M2)(R2)^{2}

I = 0.5(1.2)(0.15)^{2} + O.5 (0.45)(0.006)^{2}

I = 0.0135 + 0.0000081 = 0.0135081

ω₁ = 33 rpm = 33 x \frac{2π}{60} = 3.5 rad/s

α = -ω₁/t = -3.5 / 15 = -0.23 rad/s^{2}

torque = I x α

torque = 0.0135081 x (-0.23) = - 0.00311 N.m

torque = frictional force x R2

- 0.00311 = frictional force x 0.006

frictional force = 0.52 N

6 0

3 years ago