The White House announced Friday that a "Salute to America" will take place in the nation's capital on the Fourth of July — as the country continues to grapple with the coronavirus pandemic.
The brief bulletin indicated that the second annual event under Trump's watch is a go.
"President Donald J. Trump and First Lady Melania Trump, along with the Department of Interior, will host the 2020 Salute to America on the South Lawn of the White House and Ellipse on Saturday, July 4," the statement reads.
"In addition to music, military demonstrations, and flyovers to honor our nation's service members and veterans, the president will deliver remarks that celebrate our independence and salute our amazing heritage. The evening will culminate with a spectacular fireworks display over the National Mall."
The 2019 event was held in rainy Washington and featured tanks, fighter jets, members of the military, fireworks, and a Trump speech, among other things. It cost a reported $5.33 million.
The 2020 event will take place under the cloud of the coronavirus, which continues to ravage across the U.S. More than 121,000 Americans have died, and states continue to recommend social distancing and avoiding large crowds to help slow the spread of the virus.
In May, 10 Democrats — eight in the House, two in the Senate — from the D.C. area wrote a letter to Defense Secretary Mark Esper and Interior Secretary David Bernhardt, asking them not to allow the Fourth of July celebration to be held because of the pandemic.
The event, they claimed, "would have detrimental impacts on not only those that live in the National Capital Region, but all those who travel in from other areas of the country to attend."
"The administration, including your agencies, should be focusing on helping American families, not on a vanity project for the president."
Trump is holding a rally in Tulsa, Oklahoma, Saturday night, his first such event since the pandemic started earlier this year.
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