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In 1970, a rocket powered car called Blue Flame achieved a maximum speed of 1.00(10 km/h (278m/s).Suppose the magnitude of the c

sammy [17]

Answer:

Distance traveled during this acceleration will be 6950 m

Explanation:

Wear have given maximum speed tat will be equal to final speed of the car v = 278 m/sec

Constant acceleration $a=5.56m/sec^2$

As the car starts initially starts from rest so initial velocity of the car u = 0 m/sec

From third equation of motion $v^2=u^2+2as$

Putting all values in equation

$278^2=0^2+2\times 5.56\times s$

s = 6950 m

So distance traveled during this acceleration will be 6950 m

3 0

3 years ago

A has the magnitude 14.4 m and is angled 51.6° counterclockwise from the positive direction of the x axis of an xy coordinate sy

Ad libitum [116K]

Answer:

Ã in unit vector notation = 12.26485i + 7.54539j

B in unit vector notation = 16.3516i + 3.11529j

Explanation:

The detailed steps and calculation is shown in the attachment.

7 0

3 years ago

To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If

arlik [135]

Answer:

Part a)

$a = 1260.3 m/s^2$

Part b)

Direction = upwards

Explanation:

When ball is dropped from height h = 4.0 m

then the speed of the ball just before it will strike the ground is given as

$v_f^2 - v_i^2 = 2 a d$

$v_1^2 - 0^2 = 2(9.81)(4.0)$

$v_1 = 8.86 m/s$

Now ball will rebound to height h = 2.00 m

so the velocity of ball just after it will rebound is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - v_2^2 = 2(-9.81)(2.00)$

$v_2 = 6.26 m/s$

Part a)

Average acceleration is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{6.26 - (-8.86)}{12.0 \times 10^{-3}}$

$a = 1260.35 m/s^2$

Part B)

As we know that ball rebounds upwards after collision while before collision it is moving downwards

So the direction of the acceleration is vertically upwards

7 0

3 years ago

A diver 40 m deep in 10 degrees C fresh water exhales a 1.5 cm diameter bubble.

zzz [600]

Answer:

0.0257259766982 m

Explanation:

$P_2$ = Atmospheric pressure = 101325 Pa

$d_1$ = Initial diameter = 1.5 cm

$d_2$ = Final diameter

$\rho$ = Density of water = 1000 kg/m³

h = Depth = 40 m

The pressure is

$P_1=P_2+\rho gh\\\Rightarrow P_1=101325+1000\times 9.81\times 40\\\Rightarrow P_1=493725\ Pa$

From ideal gas law we have

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow \dfrac{P_1\dfrac{4}{3\times8}\pi d_1^3}{T_1}=\dfrac{P_2\dfrac{4}{3\times8}\pi d_2^3}{T_2}\\\Rightarrow \dfrac{P_1d_1^3}{T_1}=\dfrac{P_2d_2^3}{T_2}\\\Rightarrow d_2=(\dfrac{P_1d_1^3T_2}{P_2T_1})^{\dfrac{1}{3}}\\\Rightarrow d_2=(\dfrac{493725\times 0.015^3\times (20+273.15)}{101325\times (10+273.15)})^{\dfrac{1}{3}}\\\Rightarrow d_2=0.0257259766982\ m$

The diameter of the bubble is 0.0257259766982 m

8 0

3 years ago

Use your data (attachment) from Part 3 and Newton’s laws to explain why the force meter measures a force if the cart is moving a

vazorg [7]

The cart experiences a frictional force which is directly proportional to its weight. This means that there must be a force applied on the car to balance the forces on the car to produce a net force of 0.
This is in accordance to Newton's first law which states that an object at rest will remain at rest and an object in motion will remain in motion unless an external force acts on it. The force must be a resultant force.
Therefore, the force needed increases with the total weight of the cart as well as with the added mass in a linear manner.

4 0

3 years ago